# [LeetCode] 4Sum 四数之和

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, abcd)
• 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 ，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 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：

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());
}
};

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;
}
};

Two Sum

3Sum

4Sum II

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

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2015-05-19 23:28 Grandyang 阅读(...) 评论(...) 编辑 收藏