3Sum
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)
class Solution {
public:
vector<vector<int>> threeSum(vector<int> &num)
{
vector<vector<int>> ret;
sort(num.begin(),num.end());
for(int i=0;i<num.size();i++)
if(i>0 && num[i]==num[i-1])
continue;
else
{
int l=i+1;
int r=num.size()-1;
while(l<r)
{
if(num[l]+num[r]+num[i]==0)
{
vector<int> v;
v.push_back(num[i]);
v.push_back(num[l]);
v.push_back(num[r]);
ret.push_back(v);
l++;while(l<r && num[l]==num[l-1]) l++;
r--;while(l<r && num[r]==num[r+1]) r--;
continue;
}
if(num[l]+num[r]+num[i]>0)
{
r--;continue;
}
if(num[l]+num[r]+num[i]<0)
{
l++;continue;
}
}
}
return ret;
}
};
public:
vector<vector<int>> threeSum(vector<int> &num)
{
vector<vector<int>> ret;
sort(num.begin(),num.end());
for(int i=0;i<num.size();i++)
if(i>0 && num[i]==num[i-1])
continue;
else
{
int l=i+1;
int r=num.size()-1;
while(l<r)
{
if(num[l]+num[r]+num[i]==0)
{
vector<int> v;
v.push_back(num[i]);
v.push_back(num[l]);
v.push_back(num[r]);
ret.push_back(v);
l++;while(l<r && num[l]==num[l-1]) l++;
r--;while(l<r && num[r]==num[r+1]) r--;
continue;
}
if(num[l]+num[r]+num[i]>0)
{
r--;continue;
}
if(num[l]+num[r]+num[i]<0)
{
l++;continue;
}
}
}
return ret;
}
};

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