Now that you've solved 2sum, 3sum can be easily converted as 2sum.

Also please note the while loops for i1 and i2: they are to make sure no duplicates.

class Solution {public:    vector
     
      
        > threeSum(vector
       
         &num) {        vector
        
         
           > ret;        if (num.size() < 3) return ret;        std::sort(num.begin(), num.end());        for (int i0 = 0; i0 < num.size() - 2; i0++)        {            if (i0 > 0 && num[i0] == num[i0 - 1]) continue;            int tgt = -num[i0];            int i1 = i0 + 1, i2 = num.size() - 1;            while (i1 < i2)            {                int curr = num[i1] + num[i2];                if (curr == tgt)                {                    vector
          
            rt; rt.push_back(num[i0]); rt.push_back(num[i1]); rt.push_back(num[i2]); ret.push_back(rt); while (++i1, num[i1] == num[i1-1]); } else if (curr < tgt) { while (++i1, num[i1] == num[i1-1]); } else if (curr > tgt) { while (--i2, num[i2] == num[i2 + 1]); } } } return ret; }};