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.
把字符串T按照S的序列重排。
解决：统计T的字符，按照S的顺序依次输出。
class Solution { public: string customSortString(string S, string T) { string res; int cnt[26] = {0}; for (auto t: T) ++cnt[t-‘a‘]; for (auto s: S) { res.append(cnt[s-‘a‘], s); cnt[s-‘a‘] = 0; } for (int i=0; i<26; ++i) res.append(cnt[i], ‘a‘+i); return res; } };
