README_EN.md

15. 3Sum

中文文档

Description

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]]

Example 2:

Input: nums = []
Output: []

Example 3:

Input: nums = [0]
Output: []

 

Constraints:

• 0 <= nums.length <= 3000
• -105 <= nums[i] <= 105

Solutions

Python3

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        n, res = len(nums), []
        if n < 3:
            return res
        nums.sort()
        for i in range(n - 2):
            if nums[i] > 0:
                break
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            j, k = i + 1, n - 1
            while j < k:
                if nums[i] + nums[j] + nums[k] == 0:
                    res.append([nums[i], nums[j], nums[k]])
                    j += 1
                    k -= 1
                    while j < n and nums[j] == nums[j - 1]:
                        j += 1
                    while k > i and nums[k] == nums[k + 1]:
                        k -= 1
                elif nums[i] + nums[j] + nums[k] < 0:
                    j += 1
                else:
                    k -= 1
        return res

Java

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

C++

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int n = nums.size();
        if (n < 3) {
            return {};
        }
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        for (int i = 0; i < n - 2 && nums[i] <= 0; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int j = i + 1, k = n - 1;
            while (j < k) {
                if (nums[i] + nums[j] + nums[k] == 0) {
                    res.push_back({nums[i], nums[j], nums[k]});
                    ++j;
                    --k;
                    while (j < n && nums[j] == nums[j - 1]) ++j;
                    while (k > i && nums[k] == nums[k + 1]) --k;
                } else if (nums[i] + nums[j] + nums[k] < 0) {
                    ++j;
                } else {
                    --k;
                }
            }
        }
        return res;
    }
};

Go

func threeSum(nums []int) [][]int {
	n, res := len(nums), make([][]int, 0)
	if n < 3 {
		return res
	}
	sort.Ints(nums)
	for i := 0; i < n-2 && nums[i] <= 0; i++ {
		if i > 0 && nums[i] == nums[i-1] {
			continue
		}
		j, k := i+1, n-1
		for j < k {
			if nums[i]+nums[j]+nums[k] == 0 {
				res = append(res, []int{nums[i], nums[j], nums[k]})
				j++
				k--
				for j < n && nums[j] == nums[j-1] {
					j++
				}
				for k > i && nums[k] == nums[k+1] {
					k--
				}
			} else if nums[i]+nums[j]+nums[k] < 0 {
				j++
			} else {
				k--
			}
		}
	}
	return res
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
    const n = nums.length;
    if (n < 3) return [];
    let res = [];
    nums.sort((a, b) => a - b);
    for (let i = 0; i < n - 2 && nums[i] <= 0; ++i) {
        if (i > 0 && nums[i] == nums[i - 1]) continue;
        let j = i + 1;
        let k = n - 1;
        while (j < k) {
            if (nums[i] + nums[j] + nums[k] === 0) {
                res.push([nums[i], nums[j], nums[k]]);
                ++j;
                --k;
                while (nums[j] === nums[j - 1]) ++j;
                while (nums[k] === nums[k + 1]) --k;
            } else if (nums[i] + nums[j] + nums[k] < 0) {
                ++j;
            } else {
                --k;
            }
        }
    }
    return res;
};

...