[LeetCode] 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.

滑动子窗口(为了让代码精简，用int[128]，不要用HashMap)：

public class Solution {    public String minWindow(String s, String t) {int[] map=new int[128];int[] map2=new int[128];for(int i=0;i<t.length();i++) map[t.charAt(i)]++;int n=0;int reStart=0,reEnd=Integer.MAX_VALUE;int start=0;for(int i=0;i<s.length();i++){char ch=s.charAt(i);if(map[ch]>0){    if(map2[ch]<map[ch]) n++;map2[ch]++;}if(n>=t.length()){int k=start;for(;k<=i;k++){char c=s.charAt(k);if(map[c]==0) continue;if(map2[c]>map[c]){map2[c]--;}else break;}start=k;if(reEnd-reStart>i-start){reEnd=i;reStart=start;}}}if(reEnd==Integer.MAX_VALUE) return "";else return s.substring(reStart,reEnd+1);}}