zoukankan      html  css  js  c++  java
  • LeetCode 791. Custom Sort String

    原题链接在这里:https://leetcode.com/problems/custom-sort-string/

    题目:

    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.

    题解:

    Mark each char and frequency in T with map.

    Then for each cahr sc in S, check if it is in T as well. If yes then append it after res with frequency times. And remove it from map.

    Then the rest in map are char in T, but not in S. Append them after res.

    Time Complexity: O(m+n). m = S.length(). n = T.length().

    Space: O(n).

    AC Java:  

     1 class Solution {
     2     public String customSortString(String S, String T) {
     3         if(T == null || T.length() == 0 || S == null || S.length() == 0){
     4             return T;
     5         }
     6         
     7         int [] map = new int [26];
     8         for(char c : T.toCharArray()){
     9            map[c-'a']++;
    10         }
    11         
    12         StringBuilder sb = new StringBuilder();
    13         for(char sc : S.toCharArray()){
    14             while(map[sc-'a'] > 0){
    15                 sb.append(sc);
    16                 map[sc-'a']--;
    17             }
    18         }
    19         
    20         for(char c = 'a'; c<='z'; c++){
    21             while(map[c-'a']>0){
    22                 sb.append(c);
    23                 map[c-'a']--;
    24             }
    25         }
    26         
    27         return sb.toString();
    28     }
    29 }
  • 相关阅读:
    转】Linux下安装Tomcat服务器和部署Web应用
    转】数据描述的三个领域
    怎样在嵌入式产品中应用键值存储数据库
    图片预览插件 fancyBox
    请别昧着良心说自己的文章是原创
    学习制作操作系统 0
    HDOJ 5276 YJC tricks time multimap
    日版 Galaxy Note sc05d 涮机
    Android 自己实现 NavigationView [Design Support Library(1)]
    用标准Struts2+mvc写的用户管理
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11980686.html
Copyright © 2011-2022 走看看