文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

**解析：**Version 1，先排序，根据索引排序，对源字符串和目标字符串也根据索引的排序顺序排序，这样做主要为了判断是否有重叠字符串。遍历所有索引，如果当前索引加上源字符串的长度与下一个索引重叠，则当前索引以及下一个索引对应的字符串都不能替换。如果字符串s根据索引对应的字串与源字符串相等，则替换对应的子串，由于需要同时替换，因此s保持不变，使用result保存替换的结果，对于不进行替换的部分，根据位置保留到result中。

• Version 1

class Solution:
    def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
        k = len(indices)
        order = sorted(range(k), key=lambda x: indices[x])
        indices = [indices[i] for i in order]
        sources = [sources[i] for i in order]
        targets = [targets[i] for i in order]
        result = ''
        i = 0
        start = 0
        while i < k:
            index = indices[i]
            source = sources[i]
            target = targets[i]
            if i + 1 < k and index + len(source) - 1 >= indices[i+1]:
                i += 2
                continue
            n = len(source)
            if s[index:index+n] == source:
                result += s[start:index]
                result += target
                start = index + n
            i += 1
        result += s[start:]
        return result

• Version 2

class Solution:
    def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
        k = len(indices)
        data = sorted(zip(indices, sources, targets), key=lambda x: x[0])
        result = ''
        i = 0
        start = 0
        while i < k:
            index, source, target = data[i]
            n = len(source)
            if i + 1 < k and index + n - 1 >= data[i+1][0]:
                i += 2
                continue
            if s[index:index+n] == source:
                result += s[start:index]
                result += target
                start = index + n
            i += 1
        result += s[start:]
        return result

Reference

1. https://leetcode.com/problems/find-and-replace-in-string/