[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.

利用双指针思想，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，然后再收缩头指针，直到不能再收缩为止。

然后再向后扩展，再收缩头指针。。。

最后记录所有可能的情况中窗口最小的

public class Solution {
public String minWindow(String S, String T) {
int[] count1 = new int[256];
int[] count2 = new int[256];

for(int i = 0; i < 256; i++){
count1[i] = count2[i] = 0;
}
int count = T.length();
for(int i = 0; i < count; i++){
count1[T.charAt(i)]++;
count2[T.charAt(i)]++;
}

int start = 0;
int minSize = S.length() + 1;
int minStart = 0;

for(int i = 0; i < S.length(); i++){
if(count2[S.charAt(i)] > 0){
count1[S.charAt(i)]--;
if(count1[S.charAt(i)] >= 0)
count--;
}

if(count == 0){
while(true){
if(count2[S.charAt(start)] > 0){
if(count1[S.charAt(start)] < 0)
count1[S.charAt(start)]++;
else
break;
}
start++;
}

if(minSize > i - start + 1){
minSize = i -start + 1;
minStart = start;
}
}
}

if(minSize == S.length() + 1)
return "";
return S.substring(minStart, minStart + minSize);
}
}