[LeetCode] Minimum Window Substring 解题报告-优快云博客

本文介绍了一种在字符串S中寻找包含字符串T所有字符的最小子串的算法，采用双指针技术，通过动态维护一个区间来实现O(n)的时间复杂度。

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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.

» Solve this problem

[解题报告]
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

[Code]

1:    string minWindow(string S, string T) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:         if(S.size() == 0) return "";  
5:            if(S.size() < T.size()) return "";            
6:            int appearCount[256];  
7:            int expectCount[256];  
8:            memset(appearCount, 0, 256*sizeof(appearCount[0]));  
9:            memset(expectCount, 0, 256*sizeof(appearCount[0]));  
10:            for(int i =0; i < T.size(); i++)  
11:            {  
12:                 expectCount[T[i]]++;                 
13:            }  
14:            int minV = INT_MAX, min_start = 0;  
15:            int wid_start=0;  
16:            int appeared = 0;   
17:            for(int wid_end = 0; wid_end< S.size(); wid_end++)  
18:            {  
19:                 if(expectCount[S[wid_end]] > 0)// this char is a part of T  
20:                 {  
21:                      appearCount[S[wid_end]]++;  
22:                      if(appearCount[S[wid_end]] <= expectCount[S[wid_end]])  
23:                           appeared ++;                      
24:                 }  
25:                 if(appeared == T.size())  
26:                 {                 
27:                      while(appearCount[S[wid_start]] > expectCount[S[wid_start]]   
28:                      || expectCount[S[wid_start]] == 0)  
29:                      {  
30:                           appearCount[S[wid_start]]--;  
31:                           wid_start ++;  
32:                    }                      
33:                      if(minV > (wid_end - wid_start +1))  
34:                      {  
35:                           minV = wid_end - wid_start +1;  
36:                           min_start = wid_start;  
37:                      }  
38:                 }                 
39:            }  
40:      if(minV == INT_MAX) return "";  
41:            return S.substr(min_start, minV);      
42:    }

[已犯错误]
1. Line 8&9
不熟悉这个api，最初写成了

memset(expectCount, 0, 256);

结果老是出问题，检查了很多遍logic，也没发现有问题，最后还是放到VS下debug才发现原来是地址空间大小没有传对。正确的应该是：

memset(expectCount, 0, 256*sizeof(appearCount[0]));