[LeetCode] 15. 3Sum

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Problem

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

Note:

The solution set must not contain duplicate triplets.

Example:

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Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Explanation

  1. Sort the array.

  2. If the first element is greater than 0 or the last element is less than 0, we know that we can’t form a 0.

  3. Loop through the array start from the first element, we fix the first element, then use two pointers technique to find two elements that sum to targetVal-fixVal until break out the left < right condition. While looping, we need to ignore the same fix number and left and right values.

Solution

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class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        if (nums.length == 0 || nums[0] > 0 || nums[nums.length-1] < 0) {
            return res;
        }
        List<Integer> lst = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i-1]) {
                continue;
            }
            int fix = nums[i];
            int target = 0 - fix;
            int left = i + 1;
            int right = nums.length-1;
            while(left < right) {
                if (nums[left] + nums[right] == target) {
                    res.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    while (left < right && nums[left] == nums[left+1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right-1]) {
                        right--;
                    }
                    left++;
                    right--;
                } else if (nums[left] + nums[right] < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return res;
    }
}