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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]]
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].
Notice that the order of the output and the order of the triplets does not matter.

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 <= 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;
};

C#

public class ThreeSumComparer: IEqualityComparer<IList<int>>
{
    public bool Equals(IList<int> left, IList<int> right)
    {
        return left[0] == right[0] && left[1] == right[1] && left[2] == right[2];
    }

    public int GetHashCode(IList<int> obj)
    {
        return (obj[0] ^ obj[1] ^ obj[2]).GetHashCode();
    }
}

public class Solution {
    public IList<IList<int>> ThreeSum(int[] nums) {
        Array.Sort(nums);
        var results = new HashSet<IList<int>>(new ThreeSumComparer());

        var cIndex = Array.BinarySearch(nums, 0);
        if (cIndex < 0) cIndex = ~cIndex;
        while (cIndex < nums.Length)
        {
            var c = nums[cIndex];
            var aIndex = 0;
            var bIndex = cIndex - 1;
            while (aIndex < bIndex)
            {
                if (nums[aIndex] + nums[bIndex] + c < 0)
                {
                    var step = 1;
                    while (aIndex + step < bIndex && nums[aIndex + step] + nums[bIndex] + c < 0)
                    {
                        aIndex += step;
                        step *= 2;
                    }
                    step /= 2;
                    while (step > 0)
                    {
                        if (aIndex + step < bIndex && nums[aIndex + step] + nums[bIndex] + c < 0)
                        {
                            aIndex += step;
                        }
                        step /= 2;
                    }
                }

                if (nums[aIndex] + nums[bIndex] + c > 0)
                {
                    var step = 1;
                    while (aIndex < bIndex - step && nums[aIndex] + nums[bIndex - step] + c > 0)
                    {
                        bIndex -= step;
                        step *= 2;
                    }
                    step /= 2;
                    while (step > 0)
                    {
                        if (aIndex < bIndex - step && nums[aIndex] + nums[bIndex - step] + c > 0)
                        {
                            bIndex -= step;
                        }
                        step /= 2;
                    }
                }

                if (nums[aIndex] + nums[bIndex] + c == 0)
                {
                    var list = new List<int> { nums[aIndex], nums[bIndex], c };
                    results.Add(list);
                    ++aIndex;
                    --bIndex;
                }
                else if (nums[aIndex] + nums[bIndex] + c < 0)
                {
                    ++aIndex;
                }
                else
                {
                    --bIndex;
                }
            }
            ++cIndex;
        }

        return results.ToList();
    }
}

Ruby

# @param {Integer[]} nums
# @return {Integer[][]}
def three_sum(nums)
  res = []
  nums.sort!

  for i in 0..(nums.length - 3)
    next if i > 0 && nums[i - 1] == nums[i]
    j = i + 1
    k = nums.length - 1
    while j < k do
      sum = nums[i] + nums[j] + nums[k]
      if sum < 0
        j += 1
      elsif sum > 0
        k -= 1
      else
        res += [[nums[i], nums[j], nums[k]]]
        j += 1
        k -= 1
        j += 1 while nums[j] == nums[j - 1]
        k -= 1 while nums[k] == nums[k + 1]
      end
    end
  end

  res
end

TypeScript

function threeSum(nums: number[]): number[][] {
    nums.sort((a, b) => a - b);
    const res = [];
    const n = nums.length;
    for (let i = 0; i < n - 2; i++) {
        if (nums[i] > 0) {
            break;
        }
        const target = 0 - nums[i];
        let l = i + 1;
        let r = n - 1;
        while (l < r) {
            if (nums[l] + nums[r] === target) {
                res.push([nums[i], nums[l], nums[r]]);
                l++;
                r--;
                while (nums[l] === nums[l - 1]) {
                    l++;
                }
                while (nums[r] === nums[r + 1]) {
                    r--;
                }
            } else if (nums[l] + nums[r] < target) {
                l++;
            } else {
                r--;
            }
        }
        while (nums[i] === nums[i + 1]) {
            i++;
        }
    }
    return res;
}

Rust

use std::cmp::Ordering;

impl Solution {
    pub fn three_sum(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
        nums.sort();
        let n = nums.len();
        let mut res = vec![];
        if n < 3 {
            return res;
        }
        let mut i = 0;
        while i < n - 2 && nums[i] <= 0 {
            let mut l = i + 1;
            let mut r = n - 1;
            while l < r {
                match (nums[i] + nums[l] + nums[r]).cmp(&0) {
                    Ordering::Less => l += 1,
                    Ordering::Greater => r -= 1,
                    Ordering::Equal => {
                        res.push(vec![nums[i], nums[l], nums[r]]);
                        l += 1;
                        r -= 1;
                        while l < n && nums[l] == nums[l - 1] {
                            l += 1;
                        }
                        while r > 0 && nums[r] == nums[r + 1] {
                            r -= 1;
                        }
                    }
                }
            }
            i += 1;
            while i < n - 2 && nums[i] == nums[i - 1] {
                i += 1;
            }
        }
        res
    }
}

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