3Sum.h

	15
	Notes:
	Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0?
	Find all unique triplets in the array which gives the sum of zero.
	Note:
	Elements in a triplet (a,b,c) must be in non-descending order. (ie, a <= b <= c)
	The solution set must not contain duplicate triplets.
	For example, given array S = {-1 0 1 2 -1 -4},
	A solution set is:
	(-1, 0, 1)
	(-1, -1, 2)
	Solution: Simplify '3sum' to '2sum' O(n^2). http://en.wikipedia.org/wiki/3SUM
	*/

class Solution {

public:

vector<vector<int> > threeSum(vector<int> &num) {

vector<vector<int>> res;

sort(num.begin(), num.end());

int N = num.size();

for (int i =0; i < N-2 && num[i] <=0; ++i)

{

if (i > 0 && num[i] == num[i-1])

continue; // avoid duplicates

int twosum = 0 - num[i];

int l = i + 1, r = N - 1;

while (l < r)

{

int sum = num[l] + num[r];

if (sum < twosum)

l++;

else if (sum > twosum)

r--;

else {

res.push_back(vector<int>{num[i],num[l],num[r]});

l++; r--;

while (l < r && num[l] == num[l-1]) l++;// avoid duplicates

while (l < r && num[r] == num[r+1]) r--;// avoid duplicates

}

}

}

return res;

}

};