Minimum Window Substring

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和一个目标串T，求出在S中包含T中所有字符(包括个数)的最短子串。

O(n)

维护两个map，一个记录T中所有出现过的字符及其个数 source

另一个map记录所有T中出现过的字符，个数初始化为0  count

扫描S串

如果当前字符没有在T中出现过，跳过

如果出现过：如果当前在count中该字符的个数已经大于等于目标串中该字符的个数，则count值加1

        如果在当前count中该字符的个数小于目标串中该字符的个数，则count对应的值加1，并且判断当前的count是否都比source中的值大于或者等于。如果valid，则根据count来截取字符串：

过程为：

从left向右扫，left记录上一个不valid的位置，即初始为0

如果S[left]在T中没有出现过，跳过，left++

如果S[left]在T中出现过，则更新最短字符串, left++, count中该字符对应的值减1。当count中该字符对应的个数比source中小时，不再valid，跳出循环，从right继续向右扫描。

class Solution {public:    map<char,int> count;    map<char,int> source;    bool valid(){        for(map<char,int>::iterator it = count.begin();it!=count.end();++it){            map<char,int>::iterator it2 = source.find(it->first);            if(it->second<it2->second)                return false;        }        return true;    }    string minWindow(string S, string T) {        string mi = "";        int left = 0;        int right = 0;        int n = S.length();        for(int i=0;i<T.length();i++){            map<char,int>::iterator it = source.find(T[i]);            if(it==source.end()){                count.insert(pair<char,int>(T[i],0));                source.insert(pair<char,int>(T[i],1));            }else{                it->second++;            }        }                for(int right=0;right<n;right++){            if(source.find(S[right])!=source.end()){                map<char,int>::iterator it1 = source.find(S[right]);                    map<char,int>::iterator it2 = count.find(S[right]);                if(it1->second <= it2->second){                    it2->second++;                }else{                    it2->second++;                    if(valid()){                        while(left<=right){                            it2 = count.find(S[left]);                            if(it2==count.end()){                                left++;                                continue;                            }                            it1 = source.find(S[left]);                            if(it2->second>=it1->second){                                if(mi=="")                                    mi = S.substr(left,right-left+1);                                else                                    mi = (right-left+1)<mi.length()?S.substr(left,right-left+1):mi;                                it2->second--;                                left++;                                if(it2->second<it1->second)                                    break;                            }else{                                break;                            }                        }                    }                }            }        }        return mi;    }};

==============================================2014/10/26=====================================

以前的代码写的真是弱，能回过头来重新写一遍，而且写的更好，也是进步，鼓励一下自己

class Solution {public:    map<char,int> cnt;    string minWindow(string S, string T) {        for(int i=0;i<T.length();i++){            map<char,int>::iterator it = cnt.find(T[i]);            if(it==cnt.end())                cnt.insert(pair<char,int>(T[i],1));            else                it->second++;        }        int start = -1,length = 50000;        int m = cnt.size();        int slow = 0,fast = 0;        int n = S.length();        while(fast<n){            map<char,int>::iterator it = cnt.find(S[fast++]);            if(it!=cnt.end()){                it->second--;                if(it->second==0)                    m--;                if(m==0){                    while(slow<fast){                        it = cnt.find(S[slow]);                        if(it!=cnt.end()){                            if(fast-slow<length){                                start = slow;                                length = fast-slow;                            }                            it->second++;                            if(it->second>0){                                m++;                                slow++;                                break;                            }                        }                        slow++;                    }                }            }        }                if(start!=-1)            return S.substr(start,length);        return "";    }};