LintCode "4 Sum"
4 Pointer solution. Key: when moving pointers, we skip duplicated ones.
Ref: https://github.com/xbz/lintcode/blob/master/4_sum/4sum.cpp
class Solution {
void nextUnique(vector<int> &num, size_t &j)
{
while (j<num.size() && num[j]==num[j-]) ++j;
}
public:
vector<vector<int> > fourSum(vector<int> &num, int target)
{
vector<vector<int> > ret;
sort(num.begin(), num.end()); for (size_t i=; i<num.size(); ++i)
{
if(i > ) nextUnique(num, i); for (size_t j=i+; j<num.size(); ++j)
{
if(j>i + ) nextUnique(num, j); size_t m = j + ;
size_t n = num.size() - ;
while (m < n) {
int sum = num[i] + num[j] + num[m] + num[n];
if (sum == target)
{
vector<int> v = {num[i], num[j], num[m], num[n]};
ret.push_back(v);
m++;
n--; if(m>j+) nextUnique(num, m);
while (n<num.size()- && m<n && num[n]==num[n+]) n--;
} else if (sum < target)
++m;
else
--n;
}
}
}
return ret;
}
};
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