[LeetCode] 18. 4Sum 四数之和
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.
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
LeetCode 中关于数字之和还有其他几道,分别是 Two Sum ,3Sum ,3Sum Closest,虽然难度在递增,但是整体的套路都是一样的,在这里为了避免重复项,我们使用了 STL 中的 TreeSet,其特点是不能有重复,如果新加入的数在 TreeSet 中原本就存在的话,插入操作就会失败,这样能很好的避免的重复项的存在。此题的 O(n^3) 解法的思路跟 3Sum 基本没啥区别,就是多加了一层 for 循环,其他的都一样,代码如下:
解法一:
class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { set<vector<int>> res; sort(nums.begin(), nums.end()); for (int i = 0; i < int(nums.size() - 3); ++i) { for (int j = i + 1; j < int(nums.size() - 2); ++j) { if (j > i + 1 && nums[j] == nums[j - 1]) continue; int left = j + 1, right = nums.size() - 1; while (left < right) { long sum = (long)nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { vector<int> out{nums[i], nums[j], nums[left], nums[right]}; res.insert(out); ++left; --right; } else if (sum < target) ++left; else --right; } } } return vector<vector<int>>(res.begin(), res.end()); } };
但是毕竟用 TreeSet 来进行去重复的处理还是有些取巧,可能在 Java 中就不能这么做,那么还是来看一种比较正统的做法吧,手动进行去重复处理。主要可以进行的有三个地方,首先在两个 for 循环下可以各放一个,因为一旦当前的数字跟上面处理过的数字相同了,那么找下来肯定还是重复的。之后就是当 sum 等于 target 的时候了,在将四个数字加入结果 res 之后,left 和 right 都需要去重复处理,分别像各自的方面遍历即可,参见代码如下:
解法二:
class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { vector<vector<int>> res; int n = nums.size(); sort(nums.begin(), nums.end()); for (int i = 0; i < n - 3; ++i) { if (i > 0 && nums[i] == nums[i - 1]) continue; for (int j = i + 1; j < n - 2; ++j) { if (j > i + 1 && nums[j] == nums[j - 1]) continue; int left = j + 1, right = n - 1; while (left < right) { long sum = (long)nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { vector<int> out{nums[i], nums[j], nums[left], nums[right]}; res.push_back(out); while (left < right && nums[left] == nums[left + 1]) ++left; while (left < right && nums[right] == nums[right - 1]) --right; ++left; --right; } else if (sum < target) ++left; else --right; } } } return res; } };
Github 同步地址:
https://github.com/grandyang/leetcode/issues/18
类似题目:
参考资料:
https://leetcode.com/problems/4sum/
https://leetcode.com/problems/4sum/discuss/8549/My-16ms-c%2B%2B-code
https://leetcode.com/problems/4sum/discuss/8575/Clean-accepted-java-O(n3)-solution-based-on-3sum