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Description
Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
The solution set must not contain duplicate quadruplets.
Example:
Given array nums = [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]
]
分析
- 这里的思路也很简单，先对nums排序，然后遍历nums，找到符合条件的四个数，然后放在set集合里面，set有一个好处是，它可以去重，这样就得到了所有的答案。
- 其实就是2sum，3sum的变形，多了循环遍历而已
代码
O(n3) time complexity
O(1) space complexity
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
set<vector<int>> result;
vector<vector<int>> ans;
if(nums.size()<4){
return ans;
}
sort(nums.begin(),nums.end());
for(int i=0;i<nums.size()-3;i++){
for(int j=i+1;j<nums.size()-2;j++){
int left=j+1;
int right=nums.size()-1;
while(left<right){
int sum=nums[i]+nums[j]+nums[left]+nums[right];
if(sum==target){
vector<int> vec{nums[i],nums[j],nums[left],nums[right]};
result.insert(vec);
left++;
right--;
}else if(sum>target){
right--;
}else{
left++;
}
}
}
}
return vector<vector<int>>(result.begin(),result.end());
}
};