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。

public class Minimum_Window_Substring {
    public static String minWindow(String s, String t) {
        if(null==s||s.length()<t.length()||s.length()==0){
            return "";
        }
        
        Map<Character,Integer> map = new HashMap<>();
        int Left = 0;
        int minLeft = 0;
        int count = 0;
        int minLen = s.length()+1;
        
        for(char c:t.toCharArray()){
            if(map.containsKey(c)){
                map.put(c, map.get(c)+1);
            }else{
                map.put(c, 1);
            }
        }
        
        for(int right = 0;right<s.length();right++){
            if(map.containsKey(s.charAt(right))){
                map.put(s.charAt(right), map.get(s.charAt(right))-1);
                //统计当前包含在t中的字母,如果s.charAt(right)在t中，则count++
                if(map.get(s.charAt(right))>=0){
                    count++;
                }
                
                //当t中的元素都包含的时候
                while(count==t.length()){
                    //查看当前的(Left,right)长度是否更小
                    if(right - Left +1<minLen){
                        minLeft = Left;
                        minLen = right - Left +1;
                    }
                    
                    //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
                    if(map.containsKey(s.charAt(Left))){
                        map.put(s.charAt(Left), map.get(s.charAt(Left))+1);
                        if(map.get(s.charAt(Left))>0){
                            count--;
                        }
                    }
                    //缩小窗口左边界
                    Left++;
                }
                
            }
        }
        
        if(s.length()<minLen){
            return "";
        }
        
        return s.substring(minLeft, minLeft+minLen);
    }
    
    public static void main(String[] args) {
        System.out.println(minWindow("ADOBECODEBANC","ABC"));
    }
}