18. 4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)

The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)

和3Sum思想一样，这个复杂度是O(n^3)

class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {

vector<vector<int> > res;

if (num.empty())
return res;

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

for (int i = 0; i < num.size(); i++) {

int target_3 = target - num[i];

for (int j = i + 1; j < num.size(); j++) {

int target_2 = target_3 - num[j];

int front = j + 1;
int back = num.size() - 1;

while(front < back) {

int two_sum = num[front] + num[back];

if (two_sum < target_2) front++;

else if (two_sum > target_2) back--;

else {

vector<int> quadruplet(4, 0);
quadruplet[0] = num[i];
quadruplet[1] = num[j];
quadruplet[2] = num[front];
quadruplet[3] = num[back];
res.push_back(quadruplet);

// Processing the duplicates of number 3
while (front < back && num[front] == quadruplet[2]) ++front;

// Processing the duplicates of number 4
while (front < back && num[back] == quadruplet[3]) --back;

}
}

// Processing the duplicates of number 2
while(j + 1 < num.size() && num[j + 1] == num[j]) ++j;
}

// Processing the duplicates of number 1
while (i + 1 < num.size() && num[i + 1] == num[i]) ++i;

}

return res;

}
};