Leetcode18. 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: 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]
]

---

和之前的题目一样，不过这次我用了set容器，防止出现重复，简单一点。

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        set<vector<int>> res;
        sort(nums.begin(),nums.end());
        for(int i=0;i<int(nums.size()-3);i++)
        {
            for(int j=i+1;j<int(nums.size())-2;j++)
            {
                int l=j+1;
                int r=nums.size()-1;
                while(l<r)
                {
                    int sum=nums[i]+nums[j]+nums[l]+nums[r];
                    if(sum==target)
                    {
                        vector<int> out;
                        out.push_back(nums[i]);
                        out.push_back(nums[j]);
                        out.push_back(nums[l]);
                        out.push_back(nums[r]);
                        res.insert(out);
                        l++;
                        r--;
                    }
                    else if(sum<target)
                        l++;
                    else
                        r--;
                }
            }
        }
        return vector<vector<int>> (res.begin(),res.end());
    }
};

其中有一个问题是，在两个for循环时，如果不强制int转换的话，会超时。。。我也不是很懂为什么加上了就可以AC。。这个再研究研究~