Palindrome Partitioning
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Personally I think, the time complexity is O(n^n), n is the length of the given string.
Thank you Dan Roche, the tight time complexity = O(n* (2^n)), check details below.
#include <vector>
using namespace std;
class Solution {
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> list;
vector<string> subList;
// Input validation.
if (s.length() <= 1) {
subList.push_back(s);
list.push_back(subList);
return list;
}
int len = s.length();
vector<vector<bool>> memo(len, vector<bool>(len));
for (int i = 0; i < len; i ++) {
for (int j = 0; j < len; j ++) {
if (i >= j) memo[i][j] = true;
else memo[i][j] = false;
}
}
int start = 0;
helper(s, start, list, subList, memo);
return list;
}
void helper(string s, int start,
vector<vector<string>> &list, vector<string> &subList,
vector<vector<bool>> &memo) {
// Base case.
if (start > s.length() - 1) {
vector<string> one_rest(subList);
list.push_back(one_rest);
return;
}
for (int len = 1; start + len <= s.length(); len ++) {
int end = start + len - 1;
memo[start][end] = (len == 1) ||
(memo[start + 1][end - 1] && s[start] == s[end]);
if (memo[start][end] == true) {
// Have a try.
subList.push_back(s.substr(start, len));
// Do recursion.
helper(s, end + 1, list, subList, memo);
// Roll back.
subList.pop_back();
}
}
}
};
The worst-case running time is O(n * 2^n). This is of course exponential as you suspected, but not as bad as O(n^n).
Here's how I got O(n * 2^n): Your top-level function has an O(n^2) loop to initialize memo, and then a call to helper on the entire string. So if we write H(n) for the cost of calling helper with (s.length()-start) equal to n, then the total cost of your algorithm will be
cost = H(n) + O(n^2)
The base case for H(n) is when s.length() - start equals 1, and then it's just the cost of copying the list:
H(1) = O(n)
And for the recursive case, if the if condition memo[start][end] is true every time, there will be (n-1) recursive calls on size (n-1), (n-2), (n-3), ..., 2, 1. In addition to these recursive calls to helper, you also have to call the substr function on the same sizes, which costs O(n^2) in total. So overall the cost of H(n), for n>1, is
H(n) = H(n-1) + H(n-2) + ... + H(1) + O(n^2)
(I would write that as a summation but SO doesn't have LaTeX support.)
Now you can write the same expression for H(n-1), then substitute back to simplify:
H(n) = 2 H(n-1) + O(n)
And this solves to
H(n) = O(n * 2^n)
Since that is larger than O(n^2), the whole cost is also O(n * 2^n).
Note: You could slightly improve this by pre-computing all the substrings up front, in a single O(n^3) loop. You might as well do the same for the memo array. However, this doesn't change the asymptotic big-O bound.
In fact, O(n * 2^n) is optimal, because in the worst case the string is a repetition of the same character n times, like "aaaaaa", in which case there are 2^n possible partitions, each having size n, for a total output size of Ω(n * 2^n).
Should be O(n*2^n). You are basically trying out every possible partition out there. For a string with length n, you will have 2^(n - 1) ways to partition it. This is because, a partition is equivalent of putting a "|" in b/t two chars. There are n - 1 such slots to place a "|". There are only two choice for each slot - placing a "|" or not placing a "|". Thus 2^(n - 1) ways to place "|"s.
Then for each unique partitioning, you have to traverse the entire string (in the worst case when you have repeating chars) to make sure every partition is a palindrome. so n * 2 ^ (n - 1) = O(n*2^n).
来源：https://stackoverflow.com/questions/24591616/whats-the-time-complexity-of-this-algorithm-for-palindrome-partitioning