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.
给定一个字符串S和字符串T，找S中找一个最小的窗口，使得这个窗口内包含T中所有的字符。
1. 先建一个目标map, 记录T中字符出现的对应次数
下面以题目中所给的例子为例看一下最小窗口搜索的全过程
class Solution {
public:
string minWindow(string S, string T) {
int sizeS = S.length();
int sizeT = T.length();
if(sizeS==0 || sizeT==0 || sizeT>sizeS)return "";
//建立目标词典
int targetmap[256]; //数组中的默认值不是0，必须进行初始化
int ansmap[256];
memset(targetmap, 0, sizeof(targetmap));
memset(ansmap, 0, sizeof(ansmap));
for(int i=0; i<sizeT; i++){
targetmap[T[i]]++;
ansmap[T[i]]++;
}
//开始搜索
int minWinStart=0, minWinEnd=INT_MAX;
int p1=0, p2=0;
while(p2<sizeS){//每次一次都是找一个符合条件的窗口，然后收缩p1, 使当前窗口最小
if(targetmap[S[p2]]>0){ //只关心目标字符
ansmap[S[p2]]--;
if(ansmap[S[p2]]>=0)sizeT--; //直到T中的字符出现即可，多出现的我们不管
if(0==sizeT){ // 如果已经命中了所有的字符，则记录下当前窗口
//p1向右移动尽可能大的距离，再移动一步窗口就会被破坏
while(true){
if(targetmap[S[p1]]>0){
if(ansmap[S[p1]]<0)ansmap[S[p1]]++;
else break;
}
p1++;
}
// 记录当前这个窗口
if(p2-p1<minWinEnd-minWinStart){
minWinStart = p1;
minWinEnd = p2;
}
}
}
p2++;
}
if(minWinEnd==INT_MAX)return "";
return S.substr(minWinStart, minWinEnd-minWinStart+1);
}
};
LeetCode: Minimum Window Substring [076],布布扣,bubuko.com
LeetCode: Minimum Window Substring [076]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/27519565
