zoukankan      html  css  js  c++  java
  • [leetcode]Minimum Window Substring

    Minimum Window Substring

    Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

    For example,
    S = "ADOBECODEBANC"
    T = "ABC"

    Minimum window is "BANC".

    Note:
    If there is no such window in S that covers all characters in T, return the emtpy string "".

    If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

    算法思路:窗口移动法。 详见这里

    本题算法略吊,还是看详解吧。

    代码如下:

     1 public class Solution {
     2     public String minWindow(String s, String t) {
     3         if(s == null || t == null) return "";
     4         int[] found = new int[256];
     5         int[] needToFind = new int[256];
     6         for(int i = 0; i < t.length(); i++){
     7             needToFind[t.charAt(i)]++;
     8         }
     9         int start = 0, count = 0,minWin = Integer.MAX_VALUE;
    10         String res = "";
    11         for(int i = 0; i < s.length(); i++){
    12             if(needToFind[s.charAt(i)] == 0) continue;
    13             char c = s.charAt(i);
    14             found[c]++;
    15             if(found[c] <= needToFind[c])
    16                 count++;
    17             if(count == t.length()){
    18                 //move the begin pointer while the constrain meets
    19                 while(start < s.length() && ( needToFind[s.charAt(start)] == 0 || found[s.charAt(start)] > needToFind[s.charAt(start)])){
    20                     if(found[s.charAt(start)] > 0)
    21                         found[s.charAt(start)]--;
    22                     start++;
    23                 }
    24                 //update the min according to the current window
    25                 if(i - start + 1 < minWin){
    26                     minWin = i - start + 1;
    27                     res = s.substring(start,i + 1);
    28                 }
    29             }
    30         }
    31         return res;
    32     }
    33 }

    第二遍:

    minWin 初始化最好 > s.length(),否则当窗口大小== s.length时候,如果有重复最小窗口时候,代码可能就会稍微麻烦一点了。

    未通过case:"abc", "ac"

    这是一道很好的题目

    参考:

    http://www.cnblogs.com/jdflyfly/p/3815275.html

  • 相关阅读:
    c调用python记录
    linux 进程内存基础
    doubango类面向对象研究
    HBuilderX 连接 逍遥模拟器 之 解决没法找到模拟器设备 问题
    Application,Session,Cookie,ViewState和Cache区别
    每个.Net开发人员应该下载的十种必备工具
    ASP.NET面试资料
    .net 主题与样式
    浅谈C#当中的out关键字
    5ResponseModel响应模型
  • 原文地址:https://www.cnblogs.com/huntfor/p/3915288.html
Copyright © 2011-2022 走看看