給定一個包含 n 個整數的陣列 nums，判斷 nums 中是否存在三個元素 a，b，c ，使得 a + b + c = 0 ？找出所有滿足條件且不重複的三元組。

注意：答案中不可以包含重複的三元組。

例如, 給定陣列 nums = [-1, 0, 1, 2, -1, -4]，

滿足要求的三元組集合為：
[
  [-1, 0, 1],
  [-1, -1, 2]
]

C

/**
 * Return an array of arrays of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int** threeSum(int* nums, int numsSize, int* returnSize) 
{
    int n=numsSize;
    int** res=(int**)malloc(sizeof(int*)*(n*10));
    for(int i=0;i<(n*10);i++)
    {
        res[i]=(int*)malloc(sizeof(int)*3);
    }
    if(n<3)
    {
        *returnSize=0;
        return res;
    }
    int s=0;
    sort(nums,0,n-1);
    for(int i=0;i<=n-3;i++)
    {
        int j=i+1;
        int k=n-1;
        int m=0-nums[i];
        while(j<k)
        {
            int sum=nums[j]+nums[k];
            if(m==sum)
            {
                res[s][0]=nums[i];
                res[s][1]=nums[j++];
                res[s][2]=nums[k--];
                s++;
                while(j<k && nums[j]==nums[j-1])
                {
                    j++;
                }
                while(j<k && nums[k]==nums[k+1])
                {
                    k--;
                }
            }
            else if(sum>m)
            {
                k--;
            }
            else
            {
                j++;
            }
        }
        while(i<=n-3 && nums[i]==nums[i+1])
        {
            i++;
        }
    }
    *returnSize=s;
    return res;    
}

void sort(int *a, int left, int right)
{
    if(left >= right)
    {
        return ;
    }
    int i = left;
    int j = right;
    int key = a[i]; 
    while(i < j)                            
    {
        while(i < j && key <= a[j])
        {
            j--;
        }       
        a[i] = a[j];
        while(i < j && key >= a[i])
        {
            i++;
        }         
        a[j] = a[i];
    }    
    a[i] = key;
    sort(a, left, i - 1);
    sort(a, i + 1, right);
}
 

C++

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

python

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