[LeetCode][Java] Minimum Window Substring

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 59 阅读

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#java #数据结构与算法

本文介绍了一种在字符串S中寻找包含字符串T所有字符的最小窗口的算法，并提供了实现该算法的具体代码。采用双指针策略，尾指针先移动以包含T的所有字符，然后收缩头指针直至无法再缩小窗口。

题目：

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.

题意：

给定一个字符串S 和一个字符串T,在字符串S中找出最小的窗体，这个窗体能够包括T中的全部的字符。时间复杂度要求为O(n).

举个样例：

S = "ADOBECODEBANC"
T = "ABC"

最下的窗体为"BANC".

算法分析：

// 双指针思想，尾指针不断往后扫，当扫到有一个窗体包括了全部T的字符。然后再收缩头指针，直到不能再收缩为止。

// 最后记录全部可能的情况中窗体最小的

AC代码：

public class Solution 
{
     public String minWindow(String S, String T) 
     {
         HashMap<Character, Integer> hasFound = new HashMap<Character, Integer>();
         HashMap<Character, Integer> needToFind = new HashMap<Character, Integer>();
 
         for (int i = 0; i < T.length(); i++) 
         {
             hasFound.put(T.charAt(i), 0);
             if (needToFind.containsKey(T.charAt(i)))
             {
                 needToFind.put(T.charAt(i), needToFind.get(T.charAt(i)) + 1);
             } 
             else
             {
                 needToFind.put(T.charAt(i), 1);
             }
         }
         int begin = 0;
         int minWindowSize = S.length();
         String retString = "";
 
         int count = 0;
 
         for (int end = 0; end < S.length(); end++) 
         {
             Character end_c = S.charAt(end);
             if (needToFind.containsKey(end_c)) 
             {
                 hasFound.put(end_c, hasFound.get(end_c) + 1);
                 if (hasFound.get(end_c) <= needToFind.get(end_c)) 
                 {
                     count++;
                 }
                 if (count == T.length()) 
                 {
                    while ((!needToFind.containsKey(S.charAt(begin)))||(hasFound.get(S.charAt(begin)) > needToFind.get(S.charAt(begin)))) 
                    {
                         if (needToFind.containsKey(S.charAt(begin))) 
                         {
                             hasFound.put(S.charAt(begin),hasFound.get(S.charAt(begin)) - 1);
                         }
                         begin++;
                     }
                     if ((end - begin + 1) <= minWindowSize)
                     {
                         minWindowSize = end - begin + 1;
                         retString = S.substring(begin, end + 1);
                     }
                 }
             }
         }
         return retString;
     }
}