Minimum Window Substring
时间:2014-06-04 19:07:31
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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.
class Solution {
private:
int hashs[256];
int hasht[256];
inline bool check()
{
for(int i=0;i<256;i++)
if(hashs[i]<hasht[i]) return false;
return true;
}
public:
string minWindow(string S, string T)
{
for(int i=0;i<256;i++)
{
hashs[i]=0;
hasht[i]=0;
}
for(int i=0;i<T.size();i++)
hasht[T[i]]++;
int minl=-1;
int minr=S.length();
for(int i=0;i<S.size();i++)
hashs[S[i]]++;
if(!check()) return "";
int l=0;
int r=0;
for(int i=0;i<256;i++) hashs[i]=0;
hashs[S[0]]=1;
while(true)
{
if(check())
{
if(r-l<minr-minl)
{
minl=l;
minr=r;
if(minr-minl+1==T.length()) break;
}
hashs[S[l]]--;
l++;
}
else
{
r++;
if(r==S.length()) break;
hashs[S[r]]++;
}
}
if(minl==-1) return "";
else return S.substr(minl,minr-minl+1);
}
};
private:
int hashs[256];
int hasht[256];
inline bool check()
{
for(int i=0;i<256;i++)
if(hashs[i]<hasht[i]) return false;
return true;
}
public:
string minWindow(string S, string T)
{
for(int i=0;i<256;i++)
{
hashs[i]=0;
hasht[i]=0;
}
for(int i=0;i<T.size();i++)
hasht[T[i]]++;
int minl=-1;
int minr=S.length();
for(int i=0;i<S.size();i++)
hashs[S[i]]++;
if(!check()) return "";
int l=0;
int r=0;
for(int i=0;i<256;i++) hashs[i]=0;
hashs[S[0]]=1;
while(true)
{
if(check())
{
if(r-l<minr-minl)
{
minl=l;
minr=r;
if(minr-minl+1==T.length()) break;
}
hashs[S[l]]--;
l++;
}
else
{
r++;
if(r==S.length()) break;
hashs[S[r]]++;
}
}
if(minl==-1) return "";
else return S.substr(minl,minr-minl+1);
}
};
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