3sum.java

/*3Sum
Solved 
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] where nums[i] + nums[j] + nums[k] == 0, and the indices i, j and k are all distinct.

The output should not contain any duplicate triplets. You may return the output and the triplets in any order.

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].

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 <= 1000
-10^5 <= nums[i] <= 10^5
 */

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

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) break;
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            int l = i + 1, r = nums.length - 1;
            while (l < r) {
                int sum = nums[i] + nums[l] + nums[r];
                if (sum > 0) {
                    r--;
                } else if (sum < 0) {
                    l++;
                } else {
                    res.add(Arrays.asList(nums[i], nums[l], nums[r]));
                    l++;
                    r--;
                    while (l < r && nums[l] == nums[l - 1]) {
                        l++;
                    }
                }
            }
        }
        return res;
    }
}