LeetCode: Minimum Window Substring

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.

Solution:

这道题挺费劲的

双指针，动态维护一个区间。尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符后，然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的

用两个数组，一个保存T的中字符出现的个数expectCount['A']=1，另个保存目前为止在S中找到的T中字符的个数foundCount['A']=2

参考图解

	public String minWindow(String S, String T) {
	if(S==null||S.length()==0) return "";
	if(S.length()<T.length()) return "";
	int[] foundCount=new int[128];
	int[] expectCount=new int[128];
	int count=0;
	for(int i=0; i<T.length(); i++){
	expectCount[T.charAt(i)]++;
	}
	int minV = Integer.MAX_VALUE, minStart=0;
	int start=0, end =0;
	for(int i=0; i<S.length();i++){
	if(expectCount[S.charAt(i)]==0) continue;//skip characters not in T
	if(expectCount[S.charAt(i)]>0){//this char is part of T
	if(foundCount[S.charAt(i)] < expectCount[S.charAt(i)]){
	count++;
	}
	foundCount[S.charAt(i)]++;
	}
	if(count == T.length()){
	//from start to i, all letters in T have been found in S.substring(start, i)
	end=i;
	while(foundCount[S.charAt(start)] > expectCount[S.charAt(start)] || expectCount[S.charAt(start)]==0){
	if(foundCount[S.charAt(start)] > expectCount[S.charAt(start)])
	foundCount[S.charAt(start)]--;
	start++;
	}
	if(minV > (end-start+1)){
	minV = end-start+1;
	minStart = start;
	}
	}
	}
	if(minV == Integer.MIN_VALUE) return "";
	else return S.substring(minStart, minStart+minV);
	}

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