[leetcode] 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)

思路：我们可以仿照3sum的解决方法。这里枚举第一个和第二个数，然后对余下数的求2sum，算法复杂度为O（n^3）
参考链接：http://www.cnblogs.com/TenosDoIt/p/3649607.html

代码：

class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
int n=num.size();
vector<vector<int> > res;
sort(num.begin(),num.end());
for(int i=0;i<n-3;i++){
if(i>0 && num[i]==num[i-1]) continue;
for(int j=i+1;j<n-2;j++){
if(j>i+1 && num[j]==num[j-1]) continue;
int target2=target-num[i]-num[j];
int left=j+1,right=n-1;
while(left<right){
int temp=num[left]+num[right];
if(temp>target2) right--;
if(temp<target2) left++;
if(temp==target2){
vector<int> tempres;
tempres.push_back(num[i]);
tempres.push_back(num[j]);
tempres.push_back(num[left]);
tempres.push_back(num[right]);
res.push_back(tempres);
int k=left+1;
while(k<right && num[k]==num[left]) k++;
left=k;
k=right-1;
while(k>left && num[k]==num[right]) k--;
right=k;
}
}
}
}
return res;
}
};