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)
[Analysis]:
Brutal force. O(N^2).
[Solution]:
class Solution {
public:
vector<vector<int> > threeSum(vector<int> &num) {
vector<vector<int>> rslt;
if (num.size()<3) return rslt;
sort(num.begin(), num.end());
for (int i=0; i< num.size()-2; i++) {
if (i>0 && num[i]==num[i-1]) continue;
if (num[i]>0) break;
int l = i+1;
int r = num.size()-1;
while (l<r) {
if (num[l]+num[r]== 0-num[i]) {
rslt.push_back( vector<int>( {num[i], num[l], num[r]} ) );
l++; r--;
while (l<r && num[l-1] == num[l]) l++;
while (l<r && num[r+1]==num[r]) r--;
}
else if ( num[l]+num[r] < 0-num[i]) {
l++;
}
else {
r--;
}
}
}
return rslt;
}
};
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