leetcode-18-4sum

class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
vector<vector<int> > ret;
int len = num.size();

if (len <= 3)
return ret;

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

for (int i = 0; i <= len - 4; i++) {
for (int m = i + 1; m <= len - 3; m++) {
int j = m + 1;
int k = len - 1;
while (j < k) {
if (num[i] + num[m] + num[j] + num[k] < target) {
++j;
} else if (num[i] + num[m] +  num[j] + num[k] > target) {
--k;
} else {
ret.push_back({ num[i], num[m], num[j], num[k] });
++j;
--k;
while (j < k && num[j] == num[j - 1])
++j;
while (j < k && num[k] == num[k + 1])
--k;
}
}
while (m < len && num[m] == num[m + 1])
++m;
}
while (i < len && num[i] == num[i + 1])
++i;
}

return ret;
}
};

和3sum等思路一样。o(n*n*n)。