Minimum Window Substring 最小窗口子串问题

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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string "".
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

思路:使用到了滑动窗口，再使用Map保存目前已经匹配的目标字符串t中的元素个数。使用一个窗口(left,right)依次向右遍历。先是窗口一直向右扩张，直到(left,right)中包含目标字符串t中的所以元素的时候。

Left再向右缩，来得到最小窗口。

 1 public class Minimum_Window_Substring {
 2     public static String minWindow(String s, String t) {
 3         if(null==s||s.length()<t.length()||s.length()==0){
 4             return "";
 5         }
 6         
 7         Map<Character,Integer> map = new HashMap<>();
 8         int Left = 0;
 9         int minLeft = 0;
10         int count = 0;
11         int minLen = s.length()+1;
12         
13         for(char c:t.toCharArray()){
14             if(map.containsKey(c)){
15                 map.put(c, map.get(c)+1);
16             }else{
17                 map.put(c, 1);
18             }
19         }
20         
21         for(int right = 0;right<s.length();right++){
22             if(map.containsKey(s.charAt(right))){
23                 map.put(s.charAt(right), map.get(s.charAt(right))-1);
24                 //统计当前包含在t中的字母,如果s.charAt(right)在t中，则count++
25                 if(map.get(s.charAt(right))>=0){
26                     count++;
27                 }
28                 
29                 //当t中的元素都包含的时候
30                 while(count==t.length()){
31                     //查看当前的(Left,right)长度是否更小
32                     if(right - Left +1<minLen){
33                         minLeft = Left;
34                         minLen = right - Left +1;
35                     }
36                     
37                     //如果当前左边界包含在目标字符串t中的元素,那么则将其释放，向后遍历看是否还能匹配到更短串
38                     if(map.containsKey(s.charAt(Left))){
39                         map.put(s.charAt(Left), map.get(s.charAt(Left))+1);
40                         if(map.get(s.charAt(Left))>0){
41                             count--;
42                         }
43                     }
44                     //缩小窗口左边界
45                     Left++;
46                 }
47                 
48             }
49         }
50         
51         if(s.length()<minLen){
52             return "";
53         }
54         
55         return s.substring(minLeft, minLeft+minLen);
56     }
57     
58     public static void main(String[] args) {
59         System.out.println(minWindow("ADOBECODEBANC","ABC"));
60     }
61 }