Leetcode: 3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.
    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

Best 做法：不用Extra Space, Time Complexity still O(N^2)

Good syntax:

List<List<Integer>> res = new LinkedList<>();

res.add(Arrays.asList(num[i], num[lo], num[hi]));

 // creating Arrays of String type

 String a[] = new String[] { "A", "B", "C", "D" };

            

 // getting the list view of Array

 List<String> list = Arrays.asList(a);

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);
        
        int p = nums.length - 1;
        while (p >= 2) {
            if (p < nums.length - 1 && nums[p] == nums[p + 1]) {
                p--;
                continue;
            }
            int target = -nums[p];
            int l = 0, r = p - 1;
            while (l < r) {
                if (nums[l] + nums[r] == target) {
                    res.add(Arrays.asList(nums[l], nums[r], nums[p]));
                    while (l < r && nums[l] == nums[l + 1]) {
                        l++;
                    }
                    while (l < r && nums[r] == nums[r - 1]) {
                        r--;
                    }
                    l++;
                    r--;
                }
                else if (nums[l] + nums[r] < target) {
                    l++;
                }
                else r--;
            }
            p--;
        }
        return res;
    }
}