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)
LeetCode中關於數字之和還有其他幾道,分別是 Two Sum ,3Sum ,3Sum Closest ,雖然難度在遞增,但是整體的套路都是一樣的,在這裡為了避免重複項,我們使用了STL中的set,其特點是不能有重複,如果新加入的數在set中原本就存在的話,插入操作就會失敗,這樣能很好的避免的重複項的存在。此題的O(n^3)解法的思路跟 3Sum 基本沒啥區別,就是多加了一層for迴圈,其他的都一樣,程式碼如下:
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) { int sum = 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()); } };
但是畢竟用set來進行去重複的處理還是有些取巧,可能在Java中就不能這麼做,那麼我們還是來看一種比較正統的做法吧,手動進行去重複處理。主要可以進行的有三個地方,首先在兩個for迴圈下可以各放一個,因為一旦當前的數字跟上面處理過的數字相同了,那麼找下來肯定還是重複的。之後就是當sum等於target的時候了,我們在將四個數字加入結果res之後,left和right都需要去重複處理,分別像各自的方面遍歷即可,參見程式碼如下:
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) { int sum = 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; } };
類似題目:
參考資料:
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