15. 3Sum

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LeetCode

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Problem

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

• 3 <= nums.length <= 3000
• $-10^5 <= nums[i] <= 10^5$

Code

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums == null || nums.length < 3) return res;

        Arrays.sort(nums);

        for(int i = 0; i < nums.length - 2; i++){
            // 去重
            if(i != 0 && nums[i] == nums[i - 1]) continue;
            int left = i + 1;
            int right = nums.length - 1;
            while(left < right){
                int sum = nums[i] + nums[left] + nums[right];
                if(sum == 0){
                    res.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    // 去重
                    while(left + 1 < right && nums[left + 1] == nums[left]) left++;
                    while(right - 1 > left && nums[right - 1] == nums[right]) right--;
                    left++;
                    right--;
                } else if (sum > 0){
                    right--;
                } else {
                    left++;
                }
            }
        }

        return res;
    }
}