问题
After spending about 6-8 hours trying to digest the Manacher's algorithm, I am ready to throw in the towel. But before I do, here is one last shot in the dark: can anyone explain it? I don't care about the code. I want somebody to explain the ALGORITHM.
Here seems to be a place that others seemed to enjoy in explaining the algorithm: http://www.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html
I understand why you would want to transform the string, say, 'abba' to #a#b#b#a# After than I'm lost. For example, the author of the previously mentioned website says the key part of the algorithm is:
if P[ i' ] ≤ R – i,
then P[ i ] ← P[ i' ]
else P[ i ] ≥ P[ i' ]. (Which we have to expand past
the right edge (R) to find P[ i ])
This seems wrong, because he/she says at one point that P[i] equals 5 when P[i'] = 7 and P[i] is not less or equal to R - i.
If you are not familiar with the algorithm, here are some more links: http://tristan-interview.blogspot.com/2011/11/longest-palindrome-substring-manachers.html (I've tried this one, but the terminology is awful and confusing. First, some things are not defined. Also, too many variables. You need a checklist to recall what variable is referring to what.)
Another is: http://www.akalin.cx/longest-palindrome-linear-time (good luck)
The basic gist of the algorithm is to find the longest palindrome in linear time. It can be done in O(n^2) with a minimum to medium amount of effort. This algorithm is supposed to be quite "clever" to get it down to O(n).
回答1:
I agree that the logic isn't quite right in the explanation of the link. I give some details below.
Manacher's algorithm fills in a table P[i] which contains how far the palindrome centered at i extends. If P[5]=3, then three characters on either side of position five are part of the palindrome. The algorithm takes advantage of the fact that if you've found a long palindrome, you can fill in values of P on the right side of the palindrome quickly by looking at the values of P on the left side, since they should mostly be the same.
I'll start by explaining the case you were talking about, and then I'll expand this answer as needed.
R indicates the index of the right side of the palindrome centered at C. Here is the state at the place you indicated:
C=11
R=20
i=15
i'=7
P[i']=7
R-i=5
and the logic is like this:
if P[i']<=R-i: // not true
else: // P[i] is at least 5, but may be greater
The pseudo-code in the link indicates that P[i] should be greater than or equal to P[i'] if the test fails, but I believe it should be greater than or equal to R-i, and the explanation backs that up.
Since P[i'] is greater than R-i, the palindrome centered at i' extends past the palindrome centered at C. We know the palindrome centered at i will be at least R-i characters wide, because we still have symmetry up to that point, but we have to search explicitly beyond that.
If P[i'] had been no greater than R-i, then the largest palindrome centered at i' is within the largest palindrome centered at C, so we would have known that P[i] couldn't be any larger than P[i']. If it was, we would have a contradiction. It would mean that we would be able to extend the palindrome centered at i beyond P[i'], but if we could, then we would also be able to extend the palindrome centered at i' due to the symmetry, but it was already supposed to be as large as possible.
This case is illustrated previously:
C=11
R=20
i=13
i'=9
P[i']=1
R-i=7
In this case, P[i']<=R-i. Since we are still 7 characters away from the edge of the palindrome centered at C, we know that at least 7 characters around i are the same as the 7 characters around i'. Since there was only a one character palindrome around i', there is a one character palindrome around i as well.
j_random_hacker noticed that the logic should be more like this:
if P[i']<R-i then
P[i]=P[i']
else if P[i']>R-i then
P[i]=R-i
else P[i]=R-i + expansion
If P[i'] < R-i, then we know that P[i]==P[i'], since we're still inside the palindrome centered at C.
If P[i'] > R-i, then we know that P[i]==R-i, because otherwise the palindrome centered at C would have extended past R.
So the expansion is really only necessary in the special case where P[i']==R-i, so we don't know if the palindrome at P[i] may be longer.
This is handled in the actual code by setting P[i]=min(P[i'],R-i) and then always expanding. This way of doing it doesn't increase the time complexity, because if no expansion is necessary, the time taken to do the expansion is constant.
回答2:
The Algorithm on this site seems understandable to the certain point http://www.akalin.cx/longest-palindrome-linear-time
To understand this particular approach the best is to try to solving the problem on paper and catching the tricks you can implement to avoid checking for the palindrome for each possible center.
First answer yourself - when you find a palindrome of a given length, let's say 5 - can't you as a next step just jump to the end of this palindrome (skipping 4 letters and 4 mid-letters)?
If you try to create a palindrome with length 8 and place another palindrome with length > 8, which center is in the right side of the first palindrome you will notice something funny. Try it out: Palindrome with length 8 - WOWILIKEEKIL - Like + ekiL = 8 Now in most cases you would be able to write down the place between two E's as a center and number 8 as the length and jump after the last L to look for the center of the bigger palindrome.
This approach is not correct, which the center of bigger palindrome can be inside ekiL and you would miss it if you would jump after the last L.
After you find LIKE+EKIL you place 8 in the array that these algos use and this looks like:
[0,1,0,3,0,1,0,1,0,3,0,1,0,1,0,1,8]
for
[#,W,#,O,#,W,#,I,#,L,#,I,#,K,#,E,#]
The trick is that you already know that most probably next 7 (8-1) numbers after 8 will be the same as on the left side, so the next step is to automatically copy 7 numbers from left of 8 to right of 8 keeping in mind they are not yet final. The array would look like this
[0,1,0,3,0,1,0,1,0,3,0,1,0,1,0,1,8,1,0,1,0,1,0,3] (we are at 8)
for
[#,W,#,O,#,W,#,I,#,L,#,I,#,K,#,E,#,E,#,K,#,I,#,L]
Let's make an example, that such jump would destroy our current solution and see what we can notice.
WOWILIKEEKIL - lets try to make bigger palindrome with the center somewhere within EKIL. But its not possible - we need to change word EKIL to something that contain palindrome. What? OOOOOh - thats the trick. The only possibility to have a bigger palindrome with the center in the right side of our current palindrome is that it is already in the right (and left) side of palindrome.
Let's try to build one based on WOWILIKEEKIL We would need to change EKIL to for example EKIK with I as a center of the bigger palindrome - remember to change LIKE to KIKE as well. First letters of our tricky palindrome will be:
WOWIKIKEEKIK
as said before - let the last I be the center of the bigger pallindrome than KIKEEKIK:
WOWIKIKEEKIKEEKIKIW
let's make the array up to our old pallindrom and find out how to laverage the additional info.
for
[_ W _ O _ W _ I _ K _ I _ K _ E _ E _ K _ I _ K _ E _ E _ K _ I _ K _ I _ W ]
it will be [0,1,0,3,0,1,0,1,0,3,0,3,0,1,0,1,8
we know that the next I - a 3rd will be the longest pallindrome, but let's forget about it for a bit. lets copy the numbers in the array from the left of 8 to the right (8 numbers)
[0,1,0,3,0,1,0,1,0,3,0,3,0,1,0,1,8,1,0,1,0,3,0,3]
In our loop we are at between E's with number 8. What is special about I (future middle of biggest pallindrome) that we cannot jump right to K (the last letter of currently biggest pallindrome)? The special thing is that it exceeds the current size of the array ... how? If you move 3 spaces to the right of 3 - you are out of array. It means that it can be the middle of the biggest pallindrome and the furthest you can jump is this letter I.
Sorry for the length of this answer - I wanted to explain the algorythm and can assure you - @OmnipotentEntity was right - I understand it even better after explaining to you :)
回答3:
I have found one of the best explanation so far at the following link:
http://tarokuriyama.com/projects/palindrome2.php
It also has a visualization for the same string example (babcbabcbaccba) used at the first link mentioned in the question.
Apart from this link, i also found the code at
http://algs4.cs.princeton.edu/53substring/Manacher.java.html
I hope it will be helpful to others trying hard to understand the crux of this algorithm.
回答4:
Full Article: http://www.zrzahid.com/longest-palindromic-substring-in-linear-time-manachers-algorithm/
First of all lets observe closely to a palindrome in order to find some interesting properties. For example, S1 = "abaaba" and S2="abcba", both are palindrome but what is the non-trivial (i.e. not length or characters) difference between them? S1 is a palindrome centered around the invisible space between i=2 and i=3 (non-existent space!). On the other hand S2 is centered around character at i=2 (ie. c). In order to graciously handle the center of a palindrome irrespective of the odd/even length, lets transform the palindrome by inserting special character $ in between characters. Then S1="abba" and S2="abcba" will be transformed into T1="$a$b$a$a$b$a$" centered at i=6 and T2="$a$b$c$b$a$" centered at i=5. Now, we can see that centers are existent and lengths are consistent 2*n+1, where n=length of original string. For example,
i' c i
-----------------------------------------------------
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12|
-----------------------------------------------------
T1=| $ | a | $ | b | $ | a | $ | a | $ | b | $ | a | $ |
-----------------------------------------------------
Next, observe that from the symmetric property of a (transformed) palindrome T around the center c, T[c-k] = T[c+k] for 0<= k<= c. That is positions c-k and c+k are mirror to each other. Let's put it another way, for an index i on the right of center c, the mirror index i' is on the left of c such that c-i'=i-c => i'=2*c-i and vice versa. That is,
For each position i on the right of center c of a palindromic substring, the mirror position of i is, i'=2*c-i, and vice versa.
Let us define an array P[0..2*n] such that P[i] equals to the length of the palindrome centered at i. Note that, length is actually measured by number of characters in the original string (by ignoring special chars $). Also let min and max be respectively the leftmost and rightmost boundary of a palindromic substring centered at c. So, min=c-P[c] and max=c+P[c]. For example, for palindrome S="abaaba", the transformed palindrome T, mirror center c=6, length array P[0..12], min=c-P[c]=6-6=0, max=c+P[c]=6+6=12 and two sample mirrored indices i and i' are shown in the following figure.
min i' c i max
-----------------------------------------------------
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12|
-----------------------------------------------------
T=| $ | a | $ | b | $ | a | $ | a | $ | b | $ | a | $ |
-----------------------------------------------------
P=| 0 | 1 | 0 | 3 | 0 | 5 | 6 | 1 | 0 | 3 | 0 | 1 | 0 |
-----------------------------------------------------
With such a length array P, we can find the length of longest palindromic substring by looking into the max element of P. That is,
P[i] is the length of a palindromic substring with center at i in the transformed string T, ie. center at i/2 in the original string S; Hence the longest palindromic substring would be the substring of length P[imax] starting from index (imax-P[imax])/2 such that imax is the index of maximum element in P.
Let us draw a similar figure in the following for our non-palindromic example string S="babaabca".
min c max
|----------------|-----------------|
--------------------------------------------------------------------
idx= | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14| 15| 16|
---------------------------------------------------------------------
T=| $ | b | $ | a | $ | b | $ | a | $ | a | $ | b | $ | c | $ | a | $ |
---------------------------------------------------------------------
P=| 0 | 1 | 0 | 3 | 0 | 3 | 0 | 1 | 4 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
---------------------------------------------------------------------
Question is how to compute P efficiently. The symmetric property suggests the following cases that we could potentially use to compute P[i] by using previously computed P[i'] at the mirror index i' on the left, hence skipping a lot of computations. Let's suppose that we have a reference palindrome (first palindrome) to start with.
A third palindrome whose center is within the right side of a first palindrome will have exactly the same length as that of a second palindrome anchored at the mirror center on the left side, if the second palindrome is within the bounds of the first palindrome by at least one character.
For example in the following figure with the first palindrome centered at c=8 and bounded by min=4 and max=12, length of the third palindrome centered at i=9 (with mirror index i'= 2*c-i = 7) is, P[i] = P[i'] = 1. This is because the second palindrome centered at i' is within the bounds of first palindrome. Similarly, P[10] = P[6] = 0.|----3rd----| |----2nd----| |-----------1st Palindrome---------| min i' c i max |------------|---|---|-------------| -------------------------------------------------------------------- idx= | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14| 15| 16| --------------------------------------------------------------------- T=| $ | b | $ | a | $ | b | $ | a | $ | a | $ | b | $ | c | $ | a | $ | --------------------------------------------------------------------- P=| 0 | 1 | 0 | 3 | 0 | 3 | 0 | 1 | 4 | ? | ? | ? | ? | ? | ? | ? | ? | ---------------------------------------------------------------------Now, question is how to check this case? Note that, due to symmetric property length of segment [min..i'] is equals to the length of segment [i..max]. Also, note that 2nd palindrome is completely within 1st palindrome iff left edge of the 2nd palindrome is inside the left boundary, min of the 1st palindrome. That is,i'-P[i'] >= min =>P[i']-i' < -min (negate) =>P[i'] < i'-min =>P[i'] < max-i [(max-i)=(i'-min) due to symmetric property].Combining all the facts in case 1,P[i] = P[i'], iff (max-i) > P[i']
If the second palindrome meets or extends beyond the left bound of the first palindrome, then the third palindrome is guaranteed to have at least the length from its own center to the right outermost character of the first palindrome. This length is the same from the center of the second palindrome to the left outermost character of the first palindrome.
For example in the following figure, second palindrome centered at i=5 extends beyond the left bound of the first palindrome. So, in this case we can't say P[i]=P[i']. But length of the third palindrome centered at i=11, P[i] is at least the length from its center i=11 to the right bound max=12 of first palindrome centered at c. That is, P[i]>=1. This means third palindrome could be extended past max if and only if next immediate character past max matches exactly with the mirrored character, and we continue this check beyond. For example, in this case P[13]!=P[9] and it can't be extended. So, P[i] = 1.|-------2nd palindrome------| |----3rd----|---? |-----------1st Palindrome---------| min i' c i max |----|-----------|-----------|-----| -------------------------------------------------------------------- idx= | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14| 15| 16| --------------------------------------------------------------------- T=| $ | b | $ | a | $ | b | $ | a | $ | a | $ | b | $ | c | $ | a | $ | --------------------------------------------------------------------- P=| 0 | 1 | 0 | 3 | 0 | 3 | 0 | 1 | 4 | 1 | 0 | ? | ? | ? | ? | ? | ? | ---------------------------------------------------------------------So, how to check this case? This is simply the failed check for case 1. That is, second palindrome will extend past left edge of first palindrome iff,i'-P[i'] < min =>P[i']-i' >= -min [negate] =>P[i'] >= i'-min =>P[i'] >= max-i [(max-i)=(i'-min) due to symmetric property].That is, P[i] is at least (max-i) iff (max-i) P[i]>=(max-i), iff (max-i) Now, if the third palindrome does extend beyond max then we need to update the center and the boundary of the new palindrome.If the palindrome centered at i does expand past max then we have new (extended) palindrome, hence a new center at c=i. Update max to the rightmost boundary of the new palindrome.
Combining all the facts in case 1 and case 2, we can come up with a very beautiful little formulae:Case 1: P[i] = P[i'], iff (max-i) > P[i'] Case 2: P[i]>=(max-i), iff (max-i) = min(P[i'], max-i).That is, P[i]=min(P[i'], max-i) when the third palindrome is not extendable past max. Otherwise, we have new third palindrome at center at c=i and new max=i+P[i].
-
Neither the first nor second palindrome provides information to help determine the palindromic length of a fourth palindrome whose center is outside the right side of the first palindrome.
That is, we can't determine preemptively P[i] if i>max. That is,P[i] = 0, iff max-i < 0
Combining all the cases, we conclude the formulae:P[i] = max>i ? min(P[i'], max-i) : 0. In case we can expand beyond max then we expand by matching characters beyond max with the mirrored character with respect to new center at c=i. Finally when we have a mismatch we update new max=i+P[i].
Reference: algorithm description in wiki page
回答5:
I made a video on this Algorithm using Animation Graphics. Hopefully it will help you understand it! https://www.youtube.com/watch?v=kbUiR5YWUpQ
回答6:
This material is of great help for me to understand it: http://solutionleetcode.blogspot.com/2013/07/leetcode-longest-palindromic-substring.html
Define T as the length of the longest palindromic substrings centered at each of the characters.
The key thing is, when smaller palindromes are completely embedded within the longer palindrome, T[i] should also be symmetric within the longer palindrome.
Otherwise, we will have to compute T[i] from scratch, rather than induce from the symmetric left part.
回答7:
class Palindrome
{
private int center;
private int radius;
public Palindrome(int center, int radius)
{
if (radius < 0 || radius > center)
throw new Exception("Invalid palindrome.");
this.center = center;
this.radius = radius;
}
public int GetMirror(int index)
{
int i = 2 * center - index;
if (i < 0)
return 0;
return i;
}
public int GetCenter()
{
return center;
}
public int GetLength()
{
return 2 * radius;
}
public int GetRight()
{
return center + radius;
}
public int GetLeft()
{
return center - radius;
}
public void Expand()
{
++radius;
}
public bool LargerThan(Palindrome other)
{
if (other == null)
return false;
return (radius > other.radius);
}
}
private static string GetFormatted(string original)
{
if (original == null)
return null;
else if (original.Length == 0)
return "";
StringBuilder builder = new StringBuilder("#");
foreach (char c in original)
{
builder.Append(c);
builder.Append('#');
}
return builder.ToString();
}
private static string GetUnFormatted(string formatted)
{
if (formatted == null)
return null;
else if (formatted.Length == 0)
return "";
StringBuilder builder = new StringBuilder();
foreach (char c in formatted)
{
if (c != '#')
builder.Append(c);
}
return builder.ToString();
}
public static string FindLargestPalindrome(string str)
{
string formatted = GetFormatted(str);
if (formatted == null || formatted.Length == 0)
return formatted;
int[] radius = new int[formatted.Length];
try
{
Palindrome current = new Palindrome(0, 0);
for (int i = 0; i < formatted.Length; ++i)
{
radius[i] = (current.GetRight() > i) ?
Math.Min(current.GetRight() - i, radius[current.GetMirror(i)]) : 0;
current = new Palindrome(i, radius[i]);
while (current.GetLeft() - 1 >= 0 && current.GetRight() + 1 < formatted.Length &&
formatted[current.GetLeft() - 1] == formatted[current.GetRight() + 1])
{
current.Expand();
++radius[i];
}
}
Palindrome largest = new Palindrome(0, 0);
for (int i = 0; i < radius.Length; ++i)
{
current = new Palindrome(i, radius[i]);
if (current.LargerThan(largest))
largest = current;
}
return GetUnFormatted(formatted.Substring(largest.GetLeft(), largest.GetLength()));
}
catch (Exception ex)
{
Console.WriteLine(ex);
}
return null;
}
回答8:
Fast Javascript Solution to finding the longest palindrome in a string:
const lpal = str => {
let lpal = ""; // to store longest palindrome encountered
let pal = ""; // to store new palindromes found
let left; // to iterate through left side indices of the character considered to be center of palindrome
let right; // to iterate through left side indices of the character considered to be center of palindrome
let j; // to iterate through all characters and considering each to be center of palindrome
for (let i=0; i<str.length; i++) { // run through all characters considering them center of palindrome
pal = str[i]; // initializing current palindrome
j = i; // setting j as index at the center of palindorme
left = j-1; // taking left index of j
right = j+1; // taking right index of j
while (left >= 0 && right < str.length) { // while left and right indices exist
if(str[left] === str[right]) { //
pal = str[left] + pal + str[right];
} else {
break;
}
left--;
right++;
}
if(pal.length > lpal.length) {
lpal = pal;
}
pal = str[i];
j = i;
left = j-1;
right = j+1;
if(str[j] === str[right]) {
pal = pal + str[right];
right++;
while (left >= 0 && right < str.length) {
if(str[left] === str[right]) {
pal = str[left] + pal + str[right];
} else {
break;
}
left--;
right++;
}
if(pal.length > lpal.length) {
lpal = pal;
}
}
}
return lpal;
}
Example Input
console.log(lpal("gerngehgbrgregbeuhgurhuygbhsbjsrhfesasdfffdsajkjsrngkjbsrjgrsbjvhbvhbvhsbrfhrsbfsuhbvsuhbvhvbksbrkvkjb"));
Output
asdfffdsa
回答9:
using namespace std;
class Palindrome{
public:
Palindrome(string st){
s = st;
longest = 0;
maxDist = 0;
//ascii: 126(~) - 32 (space) = 94
// for 'a' to 'z': vector<vector<int>> v(26,vector<int>(0));
vector<vector<int>> v(94,vector<int>(0)); //all ascii
mDist.clear();
vPos = v;
bDebug = true;
};
string s;
string sPrev; //previous char
int longest; //longest palindrome size
string sLongest; //longest palindrome found so far
int maxDist; //max distance been checked
bool bDebug;
void findLongestPal();
int checkIfAnchor(int iChar, int &i);
void checkDist(int iChar, int i);
//store char positions in s pos[0] : 'a'... pos[25] : 'z'
// 0123456
// i.e. "axzebca" vPos[0][0]=0 (1st. position of 'a'), vPos[0][1]=6 (2nd pos. of 'a'),
// vPos[25][0]=2 (1st. pos. of 'z').
vector<vector<int>> vPos;
//<anchor,distance to check>
// i.e. abccba anchor = 3: position of 2nd 'c', dist = 3
// looking if next char has a dist. of 3 from previous one
// i.e. abcxcba anchor = 4: position of 2nd 'c', dist = 4
map<int,int> mDist;
};
//check if current char can be an anchor, if so return next distance to check (3 or 4)
// i.e. "abcdc" 2nd 'c' is anchor for sub-palindrome "cdc" distance = 4 if next char is 'b'
// "abcdd: 2nd 'd' is anchor for sub-palindrome "dd" distance = 3 if next char is 'c'
int Palindrome::checkIfAnchor(int iChar, int &i){
if (bDebug)
cout<<"checkIfAnchor. i:"<<i<<" iChar:"<<iChar<<endl;
int n = s.size();
int iDist = 3;
int iSize = vPos[iChar].size();
//if empty or distance to closest same char > 2
if ( iSize == 0 || vPos[iChar][iSize - 1] < (i - 2)){
if (bDebug)
cout<<" .This cannot be an anchor! i:"<<i<<" : iChar:"<<iChar<<endl;
//store char position
vPos[iChar].push_back(i);
return -1;
}
//store char position of anchor for case "cdc"
vPos[iChar].push_back(i);
if (vPos[iChar][iSize - 1] == (i - 2))
iDist = 4;
//for case "dd" check if there are more repeated chars
else {
int iRepeated = 0;
while ((i+1) < n && s[i+1] == s[i]){
i++;
iRepeated++;
iDist++;
//store char position
vPos[iChar].push_back(i);
}
}
if (bDebug)
cout<<" .iDist:"<<iDist<<" i:"<<i<<endl;
return iDist;
};
//check distance from previous same char, and update sLongest
void Palindrome::checkDist(int iChar, int i){
if (bDebug)
cout<<"CheckDist. i:"<<i<<" iChar:"<<iChar<<endl;
int iDist;
int iSize = vPos[iChar].size();
bool b1stOrLastCharInString;
bool bDiffDist;
//checkAnchor will add this char position
if ( iSize == 0){
if (bDebug)
cout<<" .1st time we see this char. Assign it INT_MAX Dist"<<endl;
iDist = INT_MAX;
}
else {
iDist = i - vPos[iChar][iSize - 1];
}
//check for distances being check, update them if found or calculate lengths if not.
if (mDist.size() == 0) {
if (bDebug)
cout<<" .no distances to check are registered, yet"<<endl;
return;
}
int i2ndMaxDist = 0;
for(auto it = mDist.begin(); it != mDist.end();){
if (bDebug)
cout<<" .mDist. anchor:"<<it->first<<" . dist:"<<it->second<<endl;
b1stOrLastCharInString = false;
bDiffDist = it->second == iDist; //check here, because it can be updated in 1st. if
if (bDiffDist){
if (bDebug)
cout<<" .Distance checked! :"<<iDist<<endl;
//check if it's the first char in the string
if (vPos[iChar][iSize - 1] == 0 || i == (s.size() - 1))
b1stOrLastCharInString = true;
//we will continue checking for more...
else {
it->second += 2; //update next distance to check
if (it->second > maxDist) {
if (bDebug)
cout<<" .previous MaxDist:"<<maxDist<<endl;
maxDist = it->second;
if (bDebug)
cout<<" .new MaxDist:"<<maxDist<<endl;
}
else if (it->second > i2ndMaxDist) {//check this...hmtest
i2ndMaxDist = it->second;
if (bDebug)
cout<<" .second MaxDist:"<<i2ndMaxDist<<endl;
}
it++;
}
}
if (!bDiffDist || b1stOrLastCharInString) {
if (bDebug && it->second != iDist)
cout<<" .Dist diff. Anchor:"<<it->first<<" dist:"<<it->second<<" iDist:"<<iDist<<endl;
else if (bDebug)
cout<<" .Palindrome found at the beggining or end of the string"<<endl;
//if we find a closest same char.
if (!b1stOrLastCharInString && it->second > iDist){
if (iSize > 1) {
if (bDebug)
cout<<" . < Dist . looking further..."<<endl;
iSize--;
iDist = i - vPos[iChar][iSize - 1];
continue;
}
}
if (iDist == maxDist) {
maxDist = 0;
if (bDebug)
cout<<" .Diff. clearing Max Dist"<<endl;
}
else if (iDist == i2ndMaxDist) {
i2ndMaxDist = 0;
if (bDebug)
cout<<" .clearing 2nd Max Dist"<<endl;
}
int iStart;
int iCurrLength;
//first char in string
if ( b1stOrLastCharInString && vPos[iChar].size() > 0 && vPos[iChar][iSize - 1] == 0){
iStart = 0;
iCurrLength = i+1;
}
//last char in string
else if (b1stOrLastCharInString && i == (s.size() - 1)){
iStart = i - it->second;
iCurrLength = it->second + 1;
}
else {
iStart = i - it->second + 1;
iCurrLength = it->second - 1; //"xabay" anchor:2nd. 'a'. Dist from 'y' to 'x':4. length 'aba':3
}
if (iCurrLength > longest){
if (bDebug)
cout<<" .previous Longest!:"<<sLongest<<" length:"<<longest<<endl;
longest = iCurrLength;
sLongest = s.substr(iStart, iCurrLength);
if (bDebug)
cout<<" .new Longest!:"<<sLongest<<" length:"<<longest<<endl;
}
if (bDebug)
cout<<" .deleting iterator for anchor:"<<it->first<<" dist:"<<it->second<<endl;
mDist.erase(it++);
}
}
//check if we need to get new max distance
if (maxDist == 0 && mDist.size() > 0){
if (bDebug)
cout<<" .new maxDist needed";
if (i2ndMaxDist > 0) {
maxDist = i2ndMaxDist;
if (bDebug)
cout<<" .assigned 2nd. max Dist to max Dist"<<endl;
}
else {
for(auto it = mDist.begin(); it != mDist.end(); it++){
if (it->second > maxDist)
maxDist = it->second;
}
if (bDebug)
cout<<" .new max dist assigned:"<<maxDist<<endl;
}
}
};
void Palindrome::findLongestPal(){
int n = s.length();
if (bDebug){
cout<<"01234567891123456789212345"<<endl<<"abcdefghijklmnopqrstuvwxyz"<<endl<<endl;
for (int i = 0; i < n;i++){
if (i%10 == 0)
cout<<i/10;
else
cout<<i;
}
cout<<endl<<s<<endl;
}
if (n == 0)
return;
//process 1st char
int j = 0;
//for 'a' to 'z' : while (j < n && (s[j] < 'a' && s[j] > 'z'))
while (j < n && (s[j] < ' ' && s[j] > '~'))
j++;
if (j > 0){
s.substr(j);
n = s.length();
}
// for 'a' to 'z' change size of vector from 94 to 26 : int iChar = s[0] - 'a';
int iChar = s[0] - ' ';
//store char position
vPos[iChar].push_back(0);
for (int i = 1; i < n; i++){
if (bDebug)
cout<<"findLongestPal. i:"<<i<<" "<<s.substr(0,i+1)<<endl;
//if max. possible palindrome would be smaller or equal
// than largest palindrome found then exit
// (n - i) = string length to check
// maxDist: max distance to check from i
int iPossibleLongestSize = maxDist + (2 * (n - i));
if ( iPossibleLongestSize <= longest){
if (bDebug)
cout<<" .PosSize:"<<iPossibleLongestSize<<" longest found:"<<iPossibleLongestSize<<endl;
return;
}
//for 'a' to 'z' : int iChar = s[i] - 'a';
int iChar = s[i] - ' ';
//for 'a' to 'z': if (iChar < 0 || iChar > 25){
if (iChar < 0 || iChar > 94){
if (bDebug)
cout<<" . i:"<<i<<" iChar:"<<s[i]<<" skipped!"<<endl;
continue;
}
//check distance to previous char, if exist
checkDist(iChar, i);
//check if this can be an anchor
int iDist = checkIfAnchor(iChar,i);
if (iDist == -1)
continue;
//append distance to check for next char.
if (bDebug)
cout<<" . Adding anchor for i:"<<i<<" dist:"<<iDist<<endl;
mDist.insert(make_pair(i,iDist));
//check if this is the only palindrome, at the end
//i.e. "......baa" or "....baca"
if (i == (s.length() - 1) && s.length() > (iDist - 2)){
//if this is the longest palindrome!
if (longest < (iDist - 1)){
sLongest = s.substr((i - iDist + 2),(iDist - 1));
}
}
}
};
int main(){
string s;
cin >> s;
Palindrome p(s);
p.findLongestPal();
cout<<p.sLongest;
return 0;
}
来源：https://stackoverflow.com/questions/10468208/manachers-algorithm-algorithm-to-find-longest-palindrome-substring-in-linear-t