LeetCode #833 - Find And Replace in String

题目描述：

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 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.

给定下标，从字符串中寻找source替换成target，由于需要从头到尾一次替换生成新的字符串，所以需要对下标排序，依次替换。

class X{
public:
    int index;
    string source;
    string target;
    X(int i, string s, string t) : index(i), source(s), target(t) {}
};

class Solution {
public:
    static bool comp(X a, X b)
    {
        if(a.index<b.index) return true;
        else return false;
    }
    
    string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
        vector<X> v;
        for(int i=0;i<indexes.size();i++)
        {
            X temp(indexes[i],sources[i],targets[i]);
            v.push_back(temp);
        }
        sort(v.begin(),v.end(),comp);
        
        string result;
        int i=0;
        int j=0;
        while(i<S.size()&&j<v.size())
        {
            if(i==v[j].index)
            {
                if(S.substr(v[j].index,v[j].source.size())==v[j].source)
                {
                    result+=v[j].target;
                    i+=v[j].source.size();
                    j++;
                }
                else
                {
                    result+=S[i];
                    i++;
                    j++;
                }
            }
            else
            {
                result+=S[i];
                i++;
            }
        }
        result+=S.substr(i);
        return result;
    }
};