[LeetCode] Minimum Window Substring

本文轉載自查看原文 2012-11-16 22:02 5597 算法/ 面試/ C++/ 模擬/ LeetCode

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.

雙指針思想，尾指針不斷往后掃，當掃到有一個窗口包含了所有T的字符，然后再收縮頭指針，直到不能再收縮為止。最后記錄所有可能的情況中窗口最小的

 1 class Solution {
 2 private:
 3     int count1[256];
 4     int count2[256];
 5 public:
 6     string minWindow(string S, string T) {
 7         // Start typing your C/C++ solution below
 8         // DO NOT write int main() function
 9         if (T.size() == 0 || S.size() == 0)
10             return "";
11             
12         memset(count1, 0, sizeof(count1));
13         memset(count2, 0, sizeof(count2));
14         
15         for(int i = 0; i < T.size(); i++)
16         {
17             count1[T[i]]++;
18             count2[T[i]]++;
19         }
20         
21         int count = T.size();
22         
23         int start = 0;
24         int minSize = INT_MAX;
25         int minStart;
26         for(int end = 0; end < S.size(); end++)
27         {
28             if (count2[S[end]] > 0)
29             {
30                 count1[S[end]]--;
31                 if (count1[S[end]] >= 0)
32                     count--;
33             }
34             
35             if (count == 0)
36             {
37                 while(true)
38                 {
39                     if (count2[S[start]] > 0)
40                     {
41                         if (count1[S[start]] < 0)
42                             count1[S[start]]++;
43                         else
44                             break;
45                     }
46                     start++;
47                 }
48             
49                 if (minSize > end - start + 1)
50                 {
51                     minSize = end - start + 1;
52                     minStart = start;
53                 }
54             }
55         }
56         
57         if (minSize == INT_MAX)
58             return "";
59         
60         string ret(S, minStart, minSize);
61         
62         return ret;        
63     }
64 };

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