Leetcode231: 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 empty string

""

.

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

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。最后记录所有可能的情况中窗口最小的

class Solution {
private:
int count1[256];
int count2[256];
public:
string minWindow(string S, string T) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (T.size() == 0 || S.size() == 0)
return "";

memset(count1, 0, sizeof(count1));
memset(count2, 0, sizeof(count2));

for(int i = 0; i < T.size(); i++)
{
count1[T[i]]++;
count2[T[i]]++;
}

int count = T.size();

int start = 0;
int minSize = INT_MAX;
int minStart;
for(int end = 0; end < S.size(); end++)
{
if (count2[S[end]] > 0)
{
count1[S[end]]--;
if (count1[S[end]] >= 0)
count--;
}
//找到了包含所有字符的end位置
if (count == 0)
{//从起点向后压缩，直到不能压缩为止
while(true)
{
if (count2[S[start]] > 0)
{
if (count1[S[start]] < 0)
count1[S[start]]++;
else
break;
}
start++;
}

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

if (minSize == INT_MAX)
return "";

string ret(S, minStart, minSize);

return ret;
}
};