Leetcode#15 3Sum

跟2sum、3sum、4sum、3sum closest一系列，参见这篇文章

排序+DFS+剪枝+二分查找

如果最后一个元素不二分查找会超时？？

代码：

 1 vector<vector<int> > res;
 2 
 3 void dfs(vector<int> &num, vector<int> ans, int pos, int left, int sum) {
 4   if (left == 0) {
 5     if (sum == 0)
 6       res.push_back(ans);
 7     return;
 8   }
 9 
10   if (left == 1) {
11     int l = pos;
12     int r = num.size() - 1;
13     while (l <= r) {
14       int m = (l + r) / 2;
15       if (num[m] + sum == 0) {
16         ans.push_back(num[m]);
17         res.push_back(ans);
18         ans.pop_back();
19         return;
20       }
21       else if (num[m] + sum < 0)
22         l = m + 1;
23       else
24         r = m - 1;
25     }
26     return;
27   }
28 
29   unordered_set<int> old;
30   for (int i = pos; i < num.size(); i++) {
31     if (old.find(num[i]) != old.end())
32       continue;
33     if (sum + left * num[i] > 0)
34       break;
35     if (sum + num[i] + (left - 1 ) * num[num.size() - 1] < 0)
36       continue;
37     old.insert(num[i]);
38     ans.push_back(num[i]);
39     dfs(num, ans, i + 1, left - 1, sum + num[i]);
40     ans.pop_back();
41   }
42 }
43 
44 vector<vector<int> > threeSum(vector<int> &num) {
45   sort(num.begin(), num.end());
46   dfs(num, vector<int>(), 0, 3, 0);
47   return res;
48 }