LeetCode in C++ 833. Find And Replace in String

To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).

Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.

For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".

Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.

All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.

Example 1:

Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:

Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:

0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters.

倒序修改，先根据indexes和sources建立对应的hashmap，因为indexes是无序的。map[i]表示S[i]起始对应的source和target。这题的难点在于字符串的各种函数应用。复习了C++ unordered_map的insert以及substr和sort（倒序），注意substr(start, length)是安全的操作，即start+length-1 out of bound时也不会报错，只返回到最后一位的substring。具体代码：

class Solution {
public:
    string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
        unordered_map<int, string> srcMap, targetMap;
        for (int i = 0; i < indexes.size(); i++) {
            srcMap.insert(pair<int, string>(indexes[i], sources[i]));
            targetMap.insert(pair<int,string>(indexes[i], targets[i]));
        }
        sort(indexes.rbegin(), indexes.rend());
        for (int i : indexes) {
            string src = srcMap.at(i);
            //C++的substr不会越界访问 以下的判断是安全的
            if (src == S.substr(i, src.size())) {
                S.replace(i, src.size(), targetMap.at(i));
            }
        }
        return S;     
    }
};

来源：https://www.cnblogs.com/lowkeysingsing/p/11291211.html

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string

replace

substr