[LeetCode] 4Sum

时间：2014-04-29 22:40:00 收藏：0 阅读：438

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

- - Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
  - The solution set must not contain duplicate quadruplets.

  For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

这题各种面熟，先来暴力的吧。因为要求abcd是递增的，所以可以先将num排序以后再爆，嗯。

我擦咧闹不住暴破法(Time Limit Exceed)：

 1     vector<vector<int> > fourSum(vector<int> &num, int target) {
 2             vector<vector<int>>result;
 3             if (num.size() < 4) return result;
 4             sort(num.begin(), num.end());
 5             for (int i=0; i<num.size()-3; i++){
 6                 for (int j=i+1; j<num.size()-2; j++){
 7                     for (int k=j+1; k<num.size()-1; k++){
 8                         for (int v=k+1; v<num.size(); v++){
 9                             if (num[i]+num[j]+num[k]+num[v] == target){
10                                 vector<int>tmp;
11                                 tmp.push_back(num[i]);
12                                 tmp.push_back(num[j]);
13                                 tmp.push_back(num[k]);
14                                 tmp.push_back(num[v]);
15                                 result.push_back(tmp);
16                             }
17                         }
18                     }
19                 }
20             }
21             return result;
22     }

O(n^4)就是个娱乐，别说大集合没戏，连稍微大一点的集合都过不了的。

看看优化，可以把最后一层循环n变成logn, 思想是最后一层循环实际上是为了从 [k+1, num.size()) 中找到一个数x且x=target - (num[i]+num[j]+num[k]) 。这样我们就可以用二分查找代替遍历，所以原本的O(n^4)就变成了O(n^3 logn)
复杂度依然是难以接受的，但至少比n^4要好一丢丢，期待进一步优化...

我擦咧稍微闹得住一丢丢暴破法(Time Limit Exceed):

 1     vector<vector<int> > fourSum(vector<int> &num, int target) {
 2             vector<vector<int>>result;
 3             if (num.size() < 4) return result;
 4             sort(num.begin(), num.end());
 5             for (int i=0; i<num.size()-3; i++){
 6                 for (int j=i+1; j<num.size()-2; j++){
 7                     for (int k=j+1; k<num.size()-1; k++){
 8                         if (binary_search(num.begin()+k+1, num.end(), target - num[i]-num[j]-num[k])){
 9                             vector<int>tmp;
10                             tmp.push_back(num[i]);
11                             tmp.push_back(num[j]);
12                             tmp.push_back(num[k]);
13                             tmp.push_back(target - num[i]-num[j]-num[k]);
14                             result.push_back(tmp);
15                         }
16                     }
17                 }
18             }
19             return result;
20     }

TO BE CONTINUE....

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