[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:

 

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());
    }
};

 

但是毕竟用set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,那么我们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,我们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) {
                    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;
    }
};

 

类似题目:

Two Sum

3Sum

4Sum II

 

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

 

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