Leetcode_002

最新推荐文章于 2022-06-01 14:01:05 发布

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博客围绕Leetcode问题展开，给定由小写字母组成的字符串S和T，S无重复字母且已按自定义顺序排序，需对T的字符重新排列，使其符合S的排序规则。给出示例及解释，还提及S和T的长度限制，最后给出C++实现。

reference:https://www.cnblogs.com/grandyang/p/9190143.html
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.

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.

implementation(C++):

class Solution {
public:
    string customSortString(string S, string T) {
        sort(T.begin(), T.end(), [&](char a, char b) {return S.find(a) < S.find(b);});
        return T;
    }
};