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minimumWindowSubstring.cpp
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minimumWindowSubstring.cpp
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// Source : https://oj.leetcode.com/problems/minimum-window-substring/
// Author : Hao Chen
// Date : 2014-07-22
/**********************************************************************************
*
* 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.
*
*
**********************************************************************************/
#include <string.h>
#include <iostream>
#include <string>
using namespace std;
#define INT_MAX 2147483647
string minWindow(string S, string T) {
string win;
if (S.size()<=0 || T.size()<=0 || T.size() > S.size()) return win;
const int MAX_CHARS = 256;
int f[MAX_CHARS], m[MAX_CHARS];
const int NOT_EXISTED = -1;
const int NOT_FOUND = 0;
memset(m, NOT_EXISTED, sizeof(m));
memset(f, NOT_EXISTED, sizeof(f));
for(int i=0; i<T.size(); i++) {
m[T[i]]==NOT_EXISTED ? m[T[i]]=1 : m[T[i]]++ ;
f[T[i]] = NOT_FOUND;
}
int start =-1;
int winSize = INT_MAX;
int letterFound = 0;
int begin = 0;
for(int i=0; i<S.size(); i++) {
if ( m[S[i]] != NOT_EXISTED ){
char ch = S[i];
f[ch]++;
if (f[ch] <= m[ch]) {
letterFound++;
}
if ( letterFound >= T.size() ) {
while ( f[S[begin]] == NOT_EXISTED || f[S[begin]] > m[S[begin]] ) {
if ( f[S[begin]] > m[S[begin]] ) {
f[S[begin]]--;
}
begin++;
}
if(winSize > i - begin + 1){
start = begin;
winSize = i - begin + 1;
}
}
}
}
if (start>=0 && winSize>0) {
win = S.substr(start, winSize);
}
return win;
}
int main(int argc, char**argv)
{
string S = "ADOBECODEBANC";
string T = "ABC";
if (argc>2){
S = argv[1];
T = argv[2];
}
cout << "S = \"" << S << "\" T=\"" << T << "\"" <<endl;
cout << minWindow(S, T) << endl;
return 0;
}