leetcode题解c++ | 76. Minimum Window Substring

题目：https://leetcode.com/problems/minimum-window-substring/#/description

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.

分析：

用双指针+map来做，我用的方法比较直观易懂。map<char, int> charactor记录t中个元素的个数，然后遍历s，遇到t中的值，就让charactor里-1，当charactor全为0时，就是一个满足条件的值。用start作左指针，当剔除start所在元素，还满足条件，且长度较小，就更新答案。

我的代码

string minWindow(string s, string t){    map<char, int> charactor;    int left=0, right=-1;    int start=0;    int num;//numbers of contained charactors    int len=INT_MAX;    for(int i=0; i<t.length(); ++i)        ++charactor[t[i]];    num = charactor.size();    for(int i=0; i<s.length(); ++i)    {        if(charactor.find(s[i]) != charactor.end())        {            --charactor[s[i]];            if(charactor[s[i]] == 0)                --num;            if(charactor[s[i]]<=0 && num==0)            {                while(start<=i)                {                    if(charactor.find(s[start])!=charactor.end() && charactor[s[start]]>=0)                    {                        if(i-start+1 < len)                        {                            left = start;                            right = i;                            len = i-start+1;                        }                        break;                    }                    if(charactor.find(s[start])!=charactor.end() && charactor[s[start]]<0)                        ++charactor[s[start]];                    ++start;                }            }        }    }    return s.substr(left, right-left+1);}

solution top1

solution里的第一个答案，有很多改进的地方，不再记录num，只用记录一个counter = t.length()，当counter=0时，就是合法值。 不需用一个真的map，可以用vector<int> map(128,0)来代替，   初始值为t中各元素的个数，移动end，所到之处-1，begin所到之处+1，这样就不需要判断begin所在的元素是不是t中，因为他们的map值永远<=0， 而在t中的元素的map值用于>=0

 top1的代码

string minWindow(string s, string t) {        vector<int> map(128,0);        for(auto c: t) map[c]++;        int counter=t.size(), begin=0, end=0, d=INT_MAX, head=0;        while(end<s.size()){            if(map[s[end++]]-->0) counter--; //in t            while(counter==0){ //valid                if(end-begin<d)  d=end-(head=begin);                if(map[s[begin++]]++==0) counter++;  //make it invalid            }          }        return d==INT_MAX? "":s.substr(head, d);    }