[Leetcode] 4Sum

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)

跟3Sum一样，先用两层循环，注意去重。

 1 class Solution {

 2 public:

 3     vector<vector<int> > fourSum(vector<int> &num, int target) {

 4         vector<vector<int> > res;

 5         vector<int> v(4);

 6         int sum;

 7         if (num.size() < 4) return res;

 8         sort(num.begin(), num.end());

 9         for (int i = 0; i < num.size() - 3; ++i) {

10             if (i > 0 && num[i] == num[i-1]) 

11                 continue;

12             for (int j = i + 1; j < num.size() - 2; ++j) {

13                 if (j > i + 1 && num[j] == num[j-1])

14                     continue;

15                 int low = j + 1, high = num.size() - 1;

16                 while (low < high) {

17                     sum = num[i] + num[j] + num[low] + num[high];

18                     if (sum > target) {

19                         --high;

20                     } else if (sum < target) {

21                         ++low;

22                     } else {

23                         v[0] = num[i];

24                         v[1] = num[j];

25                         v[2] = num[low];

26                         v[3] = num[high];

27                         res.push_back(v);

28                         while (low < num.size() && num[low+1] == num[low]) ++low;

29                         while (high > 0 && num[high-1] == num[high]) --high;

30                         ++low; --high;

31                     }

32                 }

33             }

34         }

35         return res;

36     }

37 };