791 Custom-Sort-String

difficulty: Medium

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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.

Example :
Input: 
S = "cba"
T = "abcd"
Output: "cbad"
Explanation: 
"a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a". 
Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.

Note:

• S has length at most 26, and no character is repeated in S.

• T has length at most 200.

• S and T consist of lowercase letters only.

Method One

class Solution {
    public String customSortString(String S, String T) {
        // 首先不可避免的是要统计T词频。因为很可能有重复的
        int[] count = new int[26];
        for(char c: T.toCharArray()){
            count[c-'a']++;
        }
        String ans = "";
        //然后我们把S中的都加入
        for(char c: S.toCharArray()){
            for(int i = count[c-'a']; i > 0; i--){
                ans += c;
            }
            count[c - 'a'] = 0; 
        }

        // 然后我们把剩下的再加入

        for(char c = 'a'; c <= 'z'; c++){
            for(int i = count[c-'a']; i > 0; i--){
                ans += c;
            }
        }

        return ans;
    }
}
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