4Sum

06/25/2016 Array Hash Table Two Pointers

Question

Given an array S of n integers, are there elements a, b, c, and d in S such that ? 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]
]

Solution

Result: Accepted Time: 24ms

Here should be some explanations.

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums,int target) {
        vector<vector<int>> ret;
        sort(nums.begin(), nums.end());
        const int n = nums.size();
        for (int j = 0; j < n - 3;)
        {
            for (int i =  j + 1; i < n - 2;)
            {
                int rest = target - nums[i] - nums[j],left = i + 1,right = n - 1;;
                if(rest < nums[i+1] + nums[i+2])
                    break;
                while (left < right)
                {   
                    if(rest > nums[right]+nums[right-1])
                        break;
                    if ((nums[left] + nums[right]) == rest)
                        ret.push_back(vector<int>{nums[j],nums[i],nums[left],nums[right]});
                    if ((nums[left] + nums[right]) < rest)
                        for(++left;left < right && nums[left-1] == nums[left];++left);
                    else
                        for(--right;left < right && nums[right+1] == nums[right];--right);
                }
                for(++i;nums[i-1] == nums[i];++i);
            }
            for(++j;nums[j-1] == nums[j];++j);
        }
        return ret;
    }
};

Complexity Analytics

  • Time Complexity:
  • Space Complexity: