标签：spec ase als not 字典排序 esc cdb vector ons
题目描述：
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.解题思路：
相当于桶排序，只是排序顺序不是字典排序顺序，而是自定义的顺序S。
代码：
1 class Solution { 2 public: 3 string customSortString(string S, string T) { 4 string ret; 5 vector<int> alphas(26, 0); 6 for (auto c : T) { 7 alphas[c-‘a‘]++; 8 } 9 for (auto c : S) { 10 if (alphas[c - ‘a‘] != 0) { 11 for (int i = 0; i < alphas[c - ‘a‘]; ++i) { 12 ret += c; 13 } 14 alphas[c - ‘a‘] = 0; 15 } 16 } 17 for (int i = 0; i < 26; i++) { 18 if (alphas[i] != 0) { 19 for (int j = 0; j < alphas[i]; ++j) { 20 ret += (‘a‘ + i); 21 } 22 } 23 } 24 return ret; 25 } 26 };
标签：spec ase als not 字典排序 esc cdb vector ons
原文地址：https://www.cnblogs.com/gsz-/p/9477420.html
