76. 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).

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.

重点：

• 滑动窗口
• Map的put，get，判断是否存在，遍历
• String的截取

class Solution {
    public String minWindow(String s, String t) {
        Map<Character,Integer> target = new HashMap<Character,Integer>();
        for(int i = 0; i < t.length(); i++){
            if(!target.containsKey(t.charAt(i))) target.put(t.charAt(i),1);
            else target.put(t.charAt(i), target.get(t.charAt(i))+1);
        }
        
        int left = 0;
        int right = 0;
        int minLength = Integer.MAX_VALUE;
        int minLeft = 0;
        int minRight = s.length()-1;
        Map<Character,Integer> source = new HashMap<Character,Integer>();
        source.put(s.charAt(0),1);
        while(left<=right){
            if(ifContain(source,target)){
                if(right-left+1 < minLength){
                    minLength = right-left+1;
                    minLeft = left;
                    minRight = right;
                } 
                
                source.put(s.charAt(left), source.get(s.charAt(left))-1);
                left++;
            }
            else{
                right++;
                if(right == s.length()) break;
                
                if(!source.containsKey(s.charAt(right))) source.put(s.charAt(right),1);
                else source.put(s.charAt(right), source.get(s.charAt(right))+1);
            }
        }
        
        if(minLength==Integer.MAX_VALUE) return "";
        else return s.substring(minLeft,minRight+1); 
    }
    
    public Boolean ifContain(Map<Character, Integer> source, Map<Character, Integer> target){
        for(Character key: target.keySet()){
            if(!source.containsKey(key) || source.get(key) < target.get(key)) return false;
        }
        return true;
    }
}