833. Find And Replace in String —— weekly contest 84

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.

 1 class Solution {
 2 public:
 3     string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
 4         string res;
 5         res = S;
 6         map<int,int> mp;    // indexes's value relate to index after sorting
 7         for(int i = 0; i < indexes.size();i++){
 8             mp[indexes[i]] = i;
 9         }
10         vector<int> cpi = indexes;  
11         sort(cpi.begin(),cpi.end());   //cpi.end() means the pos after the last number
12         
13         for(int i = cpi.size()-1; i>=0 ;i--){    //from big to small
14             int n = cpi[i]; 
15             int m = mp[cpi[i]];
16             if(sources[m] == S.substr(n,sources[m].size())){       //substr(int pos, int len)
17                 res.replace(n,sources[m].size(),targets[m]);       //replace(int pos, int len, string str)  or ::iterator
18             }
19         }
20         return res;
21     }
22 };

主要操作是取子串和替代的操作。

主要算法是sort和从大到小操作以避免index改变造成的不能实现同时替换的问题。

来源：https://www.cnblogs.com/jinjin-2018/p/9034076.html

标签

string

replace