76. 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 empty string "".

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

 

这种类型题目做法是创建一个target的数组，然后创建两个指针，一个start，一个end，在创建一个counter来标记是否遍历完T，此题有个难点就是考虑集中情况：

S="AAB" T="AB";S="AB" T="AB"；S="AOOB" T="AB"这三种情形。代码如下：

public class Solution {
    public String minWindow(String s, String t) {
        int[] word = new int[128];
        for(char c:t.toCharArray()){
            word[c]++;
        }
        int begin = 0;
        int end = 0;
        int count = t.length();
        int len = Integer.MAX_VALUE;
        int head = 0;
        while(end<s.length()){
            if(word[s.charAt(end++)]-->0) count--;
            while(count==0){
                if(end-begin<len) len = end-(head=begin);
                if(word[s.charAt(begin++)]++==0) count++;
            }
        }
        System.out.println(head);
        return len==Integer.MAX_VALUE?"":s.substring(head,head+len);
    }
}