473. Matchsticks to Square

灰太狼 2023-07-06 14:59 6阅读 0赞

Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used **exactly** one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

**Example 1:**

Input: [1,1,2,2,2]
    Output: true
    
    Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.

**Example 2:**

Input: [3,3,3,3,4]
    Output: false
    
    Explanation: You cannot find a way to form a square with all the matchsticks.

**Note:**

1.  The length sum of the given matchsticks is in the range of `0` to `10^9`.
2.  The length of the given matchstick array will not exceed `15`.

题意：已知一个数组，保存了n个（n <=15）火柴棍，可否使用这n个火柴棍摆成正方形。

解法一：将正方形的4条边看成4个桶，将每个火柴杆回溯的放置在每个桶中。放完n个火柴杆后，检查4个桶是否满足正方形，满足返回true，不满足返回false。在回溯的过程中，如果所有的可能都无法满足条件，则递归函数最终返回false。

![20200220120903849.png][]

优化策略：

（1）n个火柴杆的总长度和对4取余不为0时，返回假。

（2）将火柴杆从大到小排序，能减少回溯的次数。

（3）每次放置时，每条边上的的长度和小于1/4总长度。

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2xpeWF6aGVuMjAxMQ_size_16_color_FFFFFF_t_70][]

代码如下

class Solution {
    public:
       bool DFS(vector<int>& nums, int target, int i, vector<int>& res)
        {
            if (i >= nums.size())
            {
                for (int j = 0; j < 4; j++)
                {
                    if (res[j] != target)
                        return false;
                }
                return true;
            }
           
            for (int j = 0; j < 4; j++)
            {
                if (nums[i] + res[j] <= target)
                {
                    res[j] += nums[i];
                    if (DFS(nums, target, i + 1, res))               
                        return true;                
                    else                
                        res[j] -= nums[i];                
                }
            }
            return false;
        }
    
        bool makesquare(vector<int>& nums) {
            if(nums.size() < 4)
                return false;
            
            std::sort(nums.begin(), nums.end(), greater<int>());
            
            int count = 0;
            for (int i = 0; i < nums.size();i++)    
                count += nums[i];
            
            if (count % 4 != 0)        
                return false;
    
            vector<int> res(4, 0);
            return DFS(nums, count/4, 0, res);
        }   
    };

[20200220120903849.png]: https://img-blog.csdnimg.cn/20200220120903849.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2xpeWF6aGVuMjAxMQ_size_16_color_FFFFFF_t_70]: https://img-blog.csdnimg.cn/20200220132246510.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2xpeWF6aGVuMjAxMQ==,size_16,color_FFFFFF,t_70