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:
Shas length at most26, and no character is repeated inS.Thas length at most200.SandTconsist of lowercase letters only.
将T按S中出现的字母顺序排序
C++(4ms):
1 class Solution { 2 public: 3 string customSortString(string S, string T) { 4 sort(T.begin() , T.end() , [&](char a , char b){ 5 return S.find(a) < S.find(b) ; 6 }) ; 7 return T ; 8 } 9 };
本文介绍了一种基于自定义顺序的字符串排序算法实现。该算法能够根据另一字符串S中字符的出现顺序来重新排列字符串T,并提供了C++的具体实现示例。
643

被折叠的 条评论
为什么被折叠?



