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; } };