14 - 4Sum

最新推荐文章于 2024-11-28 06:15:00 发布

原创最新推荐文章于 2024-11-28 06:15:00 发布 · 551 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode Interview Questions 同时被 2 个专栏收录

45 篇文章

订阅专栏

C++

45 篇文章

订阅专栏

本文探讨了如何寻找数组中四个元素之和等于目标值的所有唯一组合。通过使用排序和双指针技巧来解决该问题，并确保解决方案集不包含重复的四元组。给出的C++实现代码详细展示了这一过程。

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)

solution: method1: 类似于3sum的方法，大概是O(n^3)时间，感觉效率有点低；想了一下没想出更优化的解法，Google了下找到一个帖子总结了k-sum问题，链接如下：

http://tech-wonderland.net/blog/summary-of-ksum-problems.html 。递归思考的话，确实k-sum问题的求解时间为O( n^(k-1) )。

method2: 还没想到，想到再更。

class Solution {
public:    
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > res;
        int size = num.size();
        
        if(size < 4)
            return res;
            
        sort(num.begin(),num.end());
        
        for(int fir = 0; fir < size - 3; fir++)
        {
            if(num[fir] == num[fir-1] && fir != 0)
                continue;
            
            for(int sec = fir +1; sec < size - 2; sec ++)
            {
                if(num[sec] == num[sec-1] && sec != fir+1)
                    continue;
                
                int start = sec + 1;
                int end = size - 1;
                while(start < end)
                {
                    int sum = num[fir] + num[sec] + num[start] + num[end];
                    if(sum == target)
                    {
                        vector<int>tuple(4,0);
                        tuple[0] = num[fir];
                        tuple[1] = num[sec];
                        tuple[2] = num[start];
                        tuple[3] = num[end];
                        
                        res.push_back(tuple);
                        start++;
                        while(num[start] == num[start-1])
                            start++;
                        end--;
                    }
                    else if(sum < target)
                    {
                        start++;
                    }
                    else 
                    {
                        end--;
                    }
                }
                
            }
        }
        
        return res;
    }
};

实现中需注意去重。