473. Matchsticks to Square

最新推荐文章于 2025-12-10 14:25:37 发布

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本文介绍了一种使用给定火柴棒长度列表构建正方形的算法。通过先判断总长度是否能被4整除，并对输入进行降序排序，采用深度优先搜索的方法尝试组合成四边等长的正方形。

problem:

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:

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

solution:

#include<algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<stdlib.h>
#include<vector>
#include<stack>

using namespace std;

class Solution {
public:
bool makesquare(vector<int>& nums) {
if (nums.empty()|| nums.size()< 4) return false;
int sum = 0;
for (int num : nums) sum += num;
if (sum % 4 != 0) return false;

sort(nums.begin(), nums.end());
reverse(nums);
int sums[4] = {0};
return dfs(nums, sums, 0, sum / 4);
}

private :
bool dfs(vector<int> nums, int sums[4], int index, int target) {
if (index == nums.size()) {
if (sums[0] == target && sums[1] == target && sums[2] == target) {
return true;
}
return false;
}

for (int i = 0; i < 4; i++) {
if (sums[i] + nums[index] > target) continue;
sums[i] += nums[index];
if (dfs(nums, sums, index + 1, target)) return true;
sums[i] -= nums[index];
}

return false;
}

private:
void reverse(vector<int>& nums) {
int i = 0, j = nums.size()- 1;
while (i < j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
i++; j--;
}
}
};
int main()
{
Solution s;
vector<int> nums;
nums.push_back(1);
nums.push_back(1);
nums.push_back(2);
nums.push_back(2);
nums.push_back(2);
cout << s.makesquare(nums) << endl;
getchar();
return 0;
}