[LeetCode]Minimum Window Substring

标签：java   leetcode   双指针   hashmap   

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.

参考：LeetCode题解 戴方勤

public class Solution {
    public String minWindow(String S, String T) {
    	if(S.length() == 0) return "";
    	Map<Character,Integer> map = new HashMap<>();
    	for(int i=0;i<T.length();i++) {
    		Integer n = map.get(T.charAt(i));
    		map.put(T.charAt(i), n==null?1:n+1);
    	}
    	Map<Character,Integer> lin = new HashMap<>();
    	int minStart = S.length();
    	int width = S.length()+1;;
    	int start = 0;
    	int app = 0;
    	for(int end = 0;end<S.length();end++){
    		char c = S.charAt(end);
    		boolean in = false;
    		if(map.containsKey(c)){
    			Integer num = lin.get(c);
    			lin.put(c, num==null?1:num+1);
    			if(lin.get(c)<=map.get(c)){
    				app++;
    			}
    		}
    		while(app==T.length()){
    			in = true;
    			char cc = S.charAt(start);
    			if(lin.containsKey(cc)){
    				lin.put(cc, lin.get(cc)-1);
    				if(lin.get(cc)<map.get(cc)){
    					app--;
    				}
    			}
    			start++;
    		}
    		if(in&&end-start+2<width){
    			minStart = start - 1;
    			width = end-start+2;
    		}
    		
    	}
    	if(width == S.length()+1) return "";
    	return S.substring(minStart,minStart+width);
    }
}

[LeetCode]Minimum Window Substring

标签：java   leetcode   双指针   hashmap   

原文地址：http://blog.csdn.net/guorudi/article/details/42914941