leetcode#791. Custom Sort String

791. Custom Sort String

Problem Description

S and T are strings composed of lowercase letters. In S, no letter occurs more than once.

S was sorted in some custom order previously. We want to permute the characters of T so that they match the order that S was sorted. More specifically, if x occurs before y in S, then x should occur before y in the returned string.

Return any permutation of T (as a string) that satisfies this property.

Solution

莫名奇妙的一道题，只要把S中每一个字符在T中出现的次数统计出来，然后重新构建一个新数组就可以了。注意把多余的字符补充到最后结果里。

class Solution {
public:
    string customSortString(string S, string T) {
        map<char, int> num;
        for (int i=0; i0;
        }
        string s;
        for (int i=0; iif (num.count(T[i]) > 0) {
                num[T[i]]++;
            } else {
                s.push_back(T[i]);
            }
        }
        for (int i=0; ifor (int j=0; jreturn s;
    }
};