Leetcode_76. Minimum Window Substring

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:s: "cbaebabacd" p: "abc"Output:[0, 6]Explanation:The substring with start index = 0 is "cba", which is an anagram of "abc".The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:s: "abab" p: "ab"Output:[0, 1, 2]Explanation:The substring with start index = 0 is "ab", which is an anagram of "ab".The substring with start index = 1 is "ba", which is an anagram of "ab".The substring with start index = 2 is "ab", which is an anagram of "ab".

题目大意:


给一个字符串S和T，找到S中最小的窗口覆盖了T中所有的字母；






解题思路：自己开始做的时候发现之前有类似的题目，但是怎么想也想不起来，很难受，类似的题目(点击打开链接)从中得到启发，其实还要细究的话我感觉是从sliding windows 之类的题目出发的。


先见一个hash表，里面藏的是T中每个字符出现的次数, 再建begin,end两个指针，用来遍历整个S字符串用的，因为这道题目要求的是最小的窗口，直接用string来记录的话不太方便，所以通过记录最小的长度来曲线救国，这样就变成了求最小窗口问题了。

代码如下:

public static String minWindow(String s, String t) {int[] hash=new int[256];for(char c:t.toCharArray())hash[c]++;int length=t.length(),begin=0,end=0,d=Integer.MAX_VALUE,head=0;while(end<s.length()){if(hash[s.charAt(end++)]-->0)length--;while(length==0){if(end-begin<d)d=end-(head=begin);if(hash[s.charAt(begin++)]++==0) length++;}}return d==Integer.MAX_VALUE?"":s.substring(head,head+d);    }

参考资料:https://leetcode.com/problems/minimum-window-substring/discuss/