zoukankan      html  css  js  c++  java
  • LeetCode 767. Reorganize String

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

    题目:

    Given a string S, check if the letters can be rearranged so that two characters that are adjacent to each other are not the same.

    If possible, output any possible result.  If not possible, return the empty string.

    Example 1:

    Input: S = "aab"
    Output: "aba"
    

    Example 2:

    Input: S = "aaab"
    Output: ""
    

    Note:

    • S will consist of lowercase letters and have length in range [1, 500].

    题解:

    In order to makie adjacent chars different from each other, we need to find the most frequent char and arrange it first.

    If the frequency of most frequent char is > (S.length()+1)/2, that means there is no way to arrange without putting 2 most frequent char adjacent, return "".

    Otherwise, arrange most prequent char first starting from index 0, index += 2.

    After this, place the rest chars starting from current index and when index hit the end, reset it to 1.

    Time Complexity: O(n). n =S.length.

    Space: O(n).

    AC Java:

     1 class Solution {
     2     public String reorganizeString(String S) {
     3         if(S == null || S.length() == 0){
     4             return S;
     5         }
     6         
     7         int n = S.length();
     8         int [] map = new int[26];
     9         int maxCount = 0;
    10         char maxChar = 'a';
    11         for(char c : S.toCharArray()){
    12             map[c - 'a']++;
    13             if(map[c - 'a'] > maxCount){
    14                 maxChar = c;
    15                 maxCount = map[c - 'a'];
    16             }
    17         }
    18         
    19         if(maxCount > (n + 1) / 2){
    20             return "";
    21         }
    22         
    23         int i = 0;
    24         char [] res = new char[n];
    25         while(map[maxChar - 'a'] > 0){
    26             res[i] = maxChar;
    27             i += 2;
    28             map[maxChar - 'a']--;
    29         }
    30         
    31         for(int k = 0; k < 26; k++){
    32             while(map[k] > 0){
    33                 if(i >= n){
    34                     i = 1;
    35                 }
    36                 
    37                 res[i] = (char)('a' + k);
    38                 i += 2;
    39                 map[k]--;
    40             }
    41         }
    42         
    43         return new String(res);
    44     }
    45 }

    类似Rearrange String k Distance Apart.

  • 相关阅读:
    分布式机器学习:算法、理论与实践——【1】
    LLVM Cookbook
    【前端】Webpack 进阶
    Noip2015 运输计划 树上差分 二分答案
    bzoj 2259: [Oibh]新型计算机 最短路 建模
    888E
    [ZJOI2012]旅游 对偶图 树的直径
    [HAOI2007]理想的正方形 单调队列 暴力
    bzoj1457: 棋盘游戏 SG函数 Nim
    Bomb HDU
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11986902.html
Copyright © 2011-2022 走看看