LeetCode | 18. 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]
]

跟2Sum 3Sum类似,用两个指针left , right 在后面遍历一遍,复杂度 O(n^3)
55ms Ac

class Solution {
public:
        vector<vector<int>> fourSum(vector<int>& nums, int target)
        {
            vector<vector<int> > res;
            vector<int> tmp;
            sort(nums.begin(),nums.end());
            int len = nums.size();
            for(int i=0;i<len;i++)
            {
                for(int j=i+1;j<len;j++)
                {
                    int left = j+1, right = len-1;
                    while(left < right)
                    {
                        int sum = nums[i]+nums[j]+nums[left]+nums[right];
                        if(sum == target)//找到目标
                        {
                            tmp.push_back(nums[i]); tmp.push_back(nums[j]);
                            tmp.push_back(nums[left]); tmp.push_back(nums[right]);
                            res.push_back(tmp);
                            while(left<len-1 && nums[left]==tmp[2])//不这样比较容易陷入死循环:因为left/right可能会没有移动
                            {
                                left++;
                            }
                            while(right>=1 && nums[right]==tmp[3])
                            {
                                right--;
                            }
                            tmp.clear();

                        }
                        else if(sum < target)
                        {
                            left++;
                        }
                        else
                        {
                            right--;
                        }
                        while(j+1<len && nums[j]==nums[j+1])
                        {
                            j++;
                        }
                    }
                }
                while(i+1<len && nums[i]==nums[i+1])
                {
                    i++;
                }
            }
            return res;
        }
};