[LeetCode]526. Beautiful Arrangement
题目描述
思路
递归
对于长度为n的数组,初始化为1,2,3,…,n的形式的数组
扫描数组,当前值满足条件时,与末位交换
然后判断n-1长度的数组的完美个数
直到n=1时,完美数组是1个,递归终止条件。
遵循的规律,每次发生交换后,数组必然为完美数组,原因?
代码
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
int countArrangement(int N) {
vector<int> nums;
for (int i = 1; i <= N; ++i)
nums.push_back(i);
return counts(N, nums);
}
int counts(int len, vector<int> &nums) {
if (len == 0) return 1;
int res = 0;
for (int i = 0; i < len; ++i) {
if (nums[i] % len == 0 || len % nums[i] == 0) {
swap(nums[i], nums[len - 1]);
res += counts(len - 1, nums);
swap(nums[i], nums[len - 1]);
}
}
return res;
}
};
int main() {
Solution s;
cout << s.countArrangement(3) << endl;
system("pause");
return 0;
}