Description

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:

  1. 0 <= indexes.length = sources.length = targets.length <= 100
  2. 0 < indexes[i] < S.length <= 1000
  3. All characters in given inputs are lowercase letters.

分析

题目的意思是：给定一个字符串，要求如果在指定位置出现了指定的soruce字符串，我们就用相应的target字符串进行替换。
-这道题目从前往后替换会很麻烦，因为替换后长度发生变化；但是从后往前替换，就不会存在这种问题了。

代码

class Solution {
public:
    string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
        vector<pair<int,int>> v;
        for(int i=0;i<indexes.size();i++){
            v.push_back(make_pair(indexes[i],i));
        }
        sort(v.rbegin(),v.rend());
        for(auto p:v){
            int i=p.first;
            string s=sources[p.second];
            string t=targets[p.second];
            if(S.substr(i,s.length())==s){
                S=S.substr(0,i)+t+S.substr(i+s.length());
            }
        }
        return S;
    }
};

参考文献

​​833. Find And Replace in String​​