LeetCode 15. 3Sums

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[1] = (-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.

public List<List<Integer>> threeSum(int[] nums) {
        
        Arrays.sort(nums);
        List<List<Integer>> res = new LinkedList<>(); 
        for (int i = 0; i < nums.length-2; i++) {
            if (i == 0 || (i > 0 && nums[i] != nums[i-1])) {
                int lo = i+1, hi = nums.length-1, sum = 0 - nums[i];
                while (lo < hi) {
                    if (nums[lo] + nums[hi] == sum) {
                        res.add(Arrays.asList(nums[i], nums[lo], nums[hi]));
                        while (lo < hi && nums[lo] == nums[lo+1]) lo++;
                        while (lo < hi && nums[hi] == nums[hi-1]) hi--;
                        lo++; hi--;
                    } else if (nums[lo] + nums[hi] < sum) lo++;
                    else hi--;
               }
            }
        }
        return res;
}