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)
Answer:
Just like "3 Sum" problem, sort the array firstly, suppose a+b+c+d = target, choose a firstly, then b secondly, then using "2 Sum" method, keep 2 pointers, in the left right elements, to find c+d == target - a -b.
Time : O(n*n*n), Space: O(1).
class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
int n=num.size();
vector<int> quadruplet(4);
vector<vector<int>> res;
if(n<4)return 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 l= j+1;
int r= n-1;
while(l < r){
if(l > j+1 && num[l]==num[l-1]) {
l++;
}
else if( r < n-1 && num[r]==num[r+1]){
r--;
}
else if(num[i]+num[j]+num[l]+num[r]< target){
l++;
}
else if(num[i]+num[j]+num[l]+num[r] > target){
r--;
}
else{
quadruplet[0]=num[i];
quadruplet[1]=num[j];
quadruplet[2]=num[l];
quadruplet[3]=num[r];
res.push_back(quadruplet);
l++;
r--;
}
}
}
}
}
};
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