转自:
http://www.cnblogs.com/remlostime/archive/2012/11/16/2774077.html
这篇文章的代码是我看到写得最简洁的。
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--;
}
if (count == 0)
{
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;
}
};