18. 4Sum

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LeetCode

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Problem

Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

• 0 <= a, b, c, d < n
• a, b, c, and d are distinct.
• nums[a] + nums[b] + nums[c] + nums[d] == target

You may return the answer in any order.

Example 1:

Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

Example 2:

Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]

Constraints:

• 1 <= nums.length <= 200
• $-10^9 <= nums[i] <= 10^9$
• $-10^9 <= target <= 10^9$

Code

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);

        for(int i = 0; i < nums.length - 3; i++){
            if(i != 0 && nums[i] == nums[i - 1]) continue;
            for(int j = i + 1; j < nums.length - 2; j++){
                if(j != i + 1 && nums[j] == nums[j - 1]) continue;

                int left = j + 1;
                int right = nums.length - 1;

                while(left < right){
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];

                    if(sum == target){
                        res.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));

                        while(left + 1 < right && nums[left + 1] == nums[left]) left++;
                        while(right - 1 > left && nums[right] == nums[right - 1]) right--;
                        left++;
                        right--;
                    } else if(sum > target){
                        right--;
                    } else if (sum < target){
                        left++;
                    }
                }
            }
        }

        return res;
    }
}