leetcode 3Sum

下面这个是笨方法。
代码：
 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 
 5 using namespace std;
 6 
 7 vector<vector<int> > threeSum(vector<int> &num) {
 8     sort(num.begin(), num.end());
 9     vector<vector<int>> result;
10     int L = num.size();
11     for (int i = 0; i < L-2; i++)
12     {
13         if (i != 0&& num[i] == num[i - 1])
14         {
15             continue;
16         }
17         for (int j = i+1; j < L-1; j++)
18         {
19             int OK = 0;
20             int left = j + 1;
21             int right = L - 1;
22             while (left <= right)
23             {
24                 int middle = (left + right) / 2;
25                 int key = 0 - num[i] - num[j];
26                 if (num[middle] == key)
27                 {
28                     vector<int> newV;
29                     newV.push_back(num[i]);
30                     newV.push_back(num[j]);
31                     newV.push_back(num[middle]);
32                     if (result.size() > 0)
33                     {
34                         if (result[result.size() - 1][0] == newV[0] && result[result.size() - 1][1] == newV[1])
35                         {
36                             break;
37                         }
38                     }
39                     result.push_back(newV);
40                     OK = 1;
41                     break;
42                 }
43                 if (OK == 1)
44                 {
45                     continue;
46                 }
47                 else
48                 {
49                     if (key < num[middle])
50                     {
51                         right = middle-1;
52                     }
53                     else
54                     {
55                         left = middle + 1;
56                     }
57                 }
58             }
59         }
60     }
61     return result;
62 }
63 
64 
65 
66 int main()
67 {
68     vector<int> S = { -1, 0, 1 ,2, -1, -4, - 1, };
69     threeSum(S);
70     return 0;
71 }
聪明的方法：