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LeetCode "3Sum"
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<vector<int> > threeSum(vector<int> &num) { vector<vector<int> > 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<int> 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; }};
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