833:Find And Replace in String

问题描述：

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:

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.

 

解析：这个题开始我想的是，直接在复制串上进行替换，但是问题出现了：怎样在动态串上找到替换的位置?于是，我想起来了使用string 里面的函数find找到子串，但是问题又出现了：有相同的子串怎样识别？  最后，发现这种方法并不可以，于是，我想起来使用静态串，标记的方式，识别完以后再进行生成目标串，这种方法证明可行，而且速度较快。

class Solution {
public:
    string findReplaceString(string S, vector& indexes, vector& sources, vector& targets) {
        string ret;
        int ind;
        int length;
        int diff = 0;
        vector  deq(S.size(),-1);
        for(int i = 0;i