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中关于数字之和还有其他几道，分别是Two Sum 两数之和，3Sum 三数之和，3Sum Closest 最近三数之和，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set，其特点是不能有重复，如果新加入的数在set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 三数之和基本没啥区别，就是多加了一层for循环，其他的都一样，代码如下：
// O(n^3) 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) { 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; out.push_back(nums[i]); out.push_back(nums[j]); out.push_back(nums[left]); out.push_back(nums[right]); res.insert(out); ++left; --right; } else if (sum < target) ++left; else --right; } } } return vector<vector<int> > (res.begin(), res.end()); } };
本文转自博客园Grandyang的博客，原文链接：四数之和[LeetCode] 4Sum ，如需转载请自行联系原博主。
