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] ]
想法：类似于3Sum的解决方式，设立双指针求解
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
int len = nums.size();
vector<vector<int>> result;
== len)
return result;
sort(nums.begin(),nums.end());
; i < len- ; i++){
&& nums.at(i) == nums.at(i-))
continue;
&& nums.at(i) + nums.at(i+)+nums.at(i+)+nums.at(i+) > target)
break;
&& nums.at(i) + nums.at(len-)+nums.at(len-)+nums.at(len-) < target)
continue;
; j < len- ; j++){
&& nums[j] == nums[j-])
continue;
&& nums[i] + nums[j] + nums[j+] + nums[j+] > target)
break;
&& nums[i] + nums[j] + nums[len-] + nums[len-] < target)
continue;
;
;
while(left < right){
int temp = nums.at(i) + nums.at(j) + nums.at(left) + nums.at(right);
if(temp == target){
result.push_back({nums.at(i),nums.at(j),nums.at(left),nums.at(right)});
left++;
right--;
])//避免重复
left++;
])
right--;
}else if(temp < target){
left++;
}else{
right--;
}
}
}
}
return result;
}
};