README_EN.md

15. 3Sum

中文文档

Description

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:

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



A solution set is:

[

  [-1, 0, 1],

  [-1, -1, 2]

]

Solutions

Python3

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        if nums is None or len(nums) < 3:
            return []
        nums.sort()
        n = len(nums)
        res = []
        for i in range(n - 2):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            p, q = i + 1, n - 1
            while p < q:
                if p > i + 1 and nums[p] == nums[p - 1]:
                    p += 1
                    continue
                if q < n - 1 and nums[q] == nums[q + 1]:
                    q -= 1
                    continue
                if nums[i] + nums[p] + nums[q] < 0:
                    p += 1
                elif nums[i] + nums[p] + nums[q] > 0:
                    q -= 1
                else:
                    res.append([nums[i], nums[p], nums[q]])
                    p += 1
                    q -= 1
        return res

Java

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        int n;
        if (nums == null || (n = nums.length) < 3) {
            return Collections.emptyList();
        }
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < n - 2; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int p = i + 1, q = n - 1;
            while (p < q) {
                if (p > i + 1 && nums[p] == nums[p - 1]) {
                    ++p;
                    continue;
                }
                if (q < n - 1 && nums[q] == nums[q + 1]) {
                    --q;
                    continue;
                }
                if (nums[p] + nums[q] + nums[i] < 0) {
                    ++p;
                } else if (nums[p] + nums[q] + nums[i] > 0) {
                    --q;
                } else {
                    res.add(Arrays.asList(nums[p], nums[q], nums[i]));
                    ++p;
                    --q;
                }
            }
        }
        return res;
    }
}

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