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]]

三个数字和为0，要求不重复。

首先想到用2Sum的方法，先确定一个数，然后内部循环使用2Sum的哈希算法。最后对结果去重。

结果超时了。贴上代码

class Solution {public:    vector
      
       
        > threeSum(vector
        
         & nums) {        vector
         
          
           > res;        if (nums.empty())            return res;        sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size(); ++i)        {            int target = 0-nums[i];            unordered_set
           
             set; vector
            
              tmp; tmp.push_back(nums[i]); for (int j = i+1; j < nums.size(); ++j) { if (set.count(target-nums[j]) > 0) { tmp.push_back(target-nums[j]); tmp.push_back(nums[j]); } if (tmp.size() == 3) { res.push_back(tmp); tmp.clear(); tmp.push_back(nums[i]); //set.clear(); } set.insert(nums[j]); } } for (auto i = res.begin(); i != res.end(); ++i) { vector
             
               vec1 = *i; sort(vec1.begin(), vec1.end()); for (auto j = i+1; j != res.end(); ++j) { vector
              
                vec2 = *j; sort(vec2.begin(), vec2.end()); if (vec1 == vec2) { j = res.erase(j); j--; } } } return res; }};
              
             
            
           
          
         
        
       
      

看了看大佬们的答案，豁然开朗

将数组排序，然后确定一个数，再使用二分查找的方法。

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