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: 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]
]
这道题是要求找到4个数之和等于给定的数字，解答方法和xxxx很像，但是复杂度要更高一些，而且这道题还有一个难点就是需要去除重复的解，这一点做的不好会导致无法AC测试用例，解答方法仍然是先排序，然后固定2个数字，用双指针遍历剩余的所有数字，4个数字之和等于target，则把这4个数作为一组解保存到结果里，4个数每一个都需要做一些措施防止重复解答
1 class Solution {
2 public:
3 vector<vector<int>> fourSum(vector<int>& nums, int target) {
4 int len = nums.size();
5 vector<vector<int>> res;
6 if (len < 4)
7 return res;
8 sort(nums.begin(), nums.end());
9 for (int i = 0; i < len - 3; i++)
10 {
11 if (i > 0 && nums[i] == nums[i-1]) continue;//处理重复
12 for (int j = i + 1; j < len - 2; j++)
13 {
14 if (j > i + 1 && nums[j] == nums[j-1]) continue;//处理重复
15 int lo = j + 1, hi = len - 1;
16 while (lo < hi)
17 {
18 int sum = nums[i] + nums[j] + nums[lo] + nums[hi];
19 if (sum == target)
20 {
21 vector<int> tmp(4, 0);
22 tmp[0] = nums[i];
23 tmp[1] = nums[j];
24 tmp[2] = nums[lo];
25 tmp[3] = nums[hi];
26 res.push_back(tmp);
27 while (lo < hi && nums[lo] == tmp[2]) lo++;//处理重复
28 while (lo < hi && nums[hi] == tmp[3]) hi--;//处理重复
29 }
30 else if (sum < target)
31 lo++;
32 else
33 hi--;
34 }
35 }
36 }
37 return res;
38 }
39 };
下面的解法很巧，利用的set元素的无重复性解答
1 // O(n^3)
2 class Solution {
3 public:
4 vector<vector<int> > fourSum(vector<int> &nums, int target) {
5 set<vector<int> > res;
6 sort(nums.begin(), nums.end());
7 for (int i = 0; i < int(nums.size() - 3); ++i) {
8 for (int j = i + 1; j < int(nums.size() - 2); ++j) {
9 int left = j + 1, right = nums.size() - 1;
10 while (left < right) {
11 int sum = nums[i] + nums[j] + nums[left] + nums[right];
12 if (sum == target) {
13 vector<int> out;
14 out.push_back(nums[i]);
15 out.push_back(nums[j]);
16 out.push_back(nums[left]);
17 out.push_back(nums[right]);
18 res.insert(out);
19 ++left; --right;
20 } else if (sum < target) ++left;
21 else --right;
22 }
23 }
24 }
25 return vector<vector<int> > (res.begin(), res.end());
26 }
27 };
