source: https://leetcode.com/problems/4sum/
C/C++ Solution to LeetCode problem 18. 4Sum.
Problem
Given an array nums
of n
integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]]
such that:
0 <= a, b, c, d < n
a
,b
,c
, andd
are distinct.nums[a] + nums[b] + nums[c] + nums[d] == target
You may return the answer in any order.
Examples
Example 1:
Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Example 2:
Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]
Constraints
1 <= nums.length <= 200
-109 <= nums[i] <= 109
-109 <= target <= 109
Solution
This is an extension of the 3Sum problem, that is an extension of the 2Sum problem.
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class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
if (nums.size() < 4)
return {};
vector<vector<int>> result;
sort(nums.begin(), nums.end());
for (int i=0; i<nums.size() - 3; i+=1) {
if (i>0 && nums[i] == nums[i-1])
continue;
for (int j=i+1; j<nums.size() - 2; j+=1) {
if (j>i+1 && nums[j] == nums[j-1])
continue;
int l = j+1;
int r = nums.size() - 1;
long sum = nums[i] + nums[j];
while (l < r) {
long s = sum + nums[l] + nums[r];
if (s == target) {
result.push_back({nums[i], nums[j], nums[l], nums[r]});
l += 1;
r -= 1;
while (l < r && nums[l] == nums[l-1])
l += 1;
while (l < r && nums[r] == nums[r+1])
r -= 1;
}
else if (s > target)
r -= 1;
else
l += 1;
}
}
}
return result;
}
};