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.

 

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思路：

双指针，动态维护一个区间。尾指针不断往后扫，当扫到有一个窗口包含了所有 T 的字符后，

然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的

时间复杂度 O(n)，空间复杂度 O(1)

class Solution {    public:        string minWindow(string S, string T)        {            if(S.empty() || T.empty() || S.size() < T.size())                return string();            vector
     
       expect(256, 0);            vector
      
        appear(256, 0);            for(int i = 0; i < T.size(); i++)            {                   expect[T[i]] ++;             }               int minWidth = INT_MAX, min_start = 0; //            int win_start = 0;            int appearCharCnt = 0;            for(int win_end = 0; win_end < S.size(); win_end++)            {                   if(expect[S[win_end]] > 0) // this char is part of T                {                       appear[S[win_end]]++;                    if(appear[S[win_end]] <= expect[S[win_end]])                        appearCharCnt ++;                 }                   //cout << "appearCharCnt\t" <
       
        << endl;                if(appearCharCnt == T.size())                {                       // shrink the start                    while (appear[S[win_start]] > expect[S[win_start]]                            || expect[S[win_start]] == 0) {                        appear[S[win_start]]--;                        win_start++;                    }                    if ((win_end - win_start + 1) < minWidth) {                        minWidth = win_end - win_start + 1;                        min_start = win_start;                        //cout << "min_start\t" <
        
         << endl;                        //cout << "min_width\t" <
         
          << endl;                    }                }            }            if (minWidth == INT_MAX)                return "";            else                return S.substr(min_start, minWidth);        }};
         
        
       
      
     

 

精简一下条件判断

class Solution {    public:        string minWindow(string S, string T)        {            if(S.empty() || T.empty() || S.size() < T.size())                return string();            vector
     
       expect(256, 0);            vector
      
        appear(256, 0);            for(int i = 0; i < T.size(); i++)            {                expect[T[i]] ++;            }            int minWidth = INT_MAX, min_start = 0;             int win_start = 0;            int appearCharCnt = 0;            for(int win_end = 0; win_end < S.size(); win_end++)            {                   appear[S[win_end]]++;                if(appear[S[win_end]] <= expect[S[win_end]])                    appearCharCnt ++;                 //cout << "appearCharCnt\t" <
       
        << endl;                if(appearCharCnt == T.size())                {                       // shrink the win start                    while (appear[S[win_start]] > expect[S[win_start]]                          ) {                         appear[S[win_start]]--;                        win_start++;                    }                       if ((win_end - win_start + 1) < minWidth) {                        minWidth = win_end - win_start + 1;                        min_start = win_start;                        //cout << "min_start\t" <
        
         << endl;                        //cout << "min_width\t" <
         
          << endl;                    }                }            }            if (minWidth == INT_MAX)                return "";            else                return S.substr(min_start, minWidth);        }};