[Leetcode] Minimum Window Substring (Java)

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.

在S中找到包含T所有字符的最小窗口（类似于文本摘要提取）

用两个指针：start、end

1)若未包含所有字符，则end++；

2)若包含所有字符，则start++，直到不包含为止，得到窗口大小并与当前最小窗口比较并更新。

public class Solution {    public String minWindow(String S, String T) {int min = S.length()+1;int minStart = 0;boolean flag = false;int count=T.length();int[] count1 = new int[256];Map<Character, Boolean> map = new HashMap<Character, Boolean>();int start=0,end=0;for(int i=0;i<T.length();i++){count1[T.charAt(i)]++;map.put(T.charAt(i), true);}for(end=0;end<S.length();end++){if(map.containsKey(S.charAt(end))&&map.get(S.charAt(end))){count1[S.charAt(end)]--;if(count1[S.charAt(end)]>=0){count--;}}if(count==0){while(count==0){if(map.containsKey(S.charAt(start))&&map.get(S.charAt(start))){count1[S.charAt(start)]++;if(count1[S.charAt(start)]>0)count++;}start++;}if(end-start+2<min){min=end-start+2;minStart=start-1;}}}if(min==S.length()+1)return "";    return S.substring(minStart,minStart+min);    }}