0526. Beautiful Arrangement

526. Beautiful Arrangement #

题目 #

Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:

  1. The number at the i position is divisible by i.th
  2. i is divisible by the number at the i position.th

Now given N, how many beautiful arrangements can you construct?

Example 1:

Input: 2
Output: 2
Explanation: 

The first beautiful arrangement is [1, 2]:

Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).

Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).

The second beautiful arrangement is [2, 1]:

Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).

Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.

Note:

  1. N is a positive integer and will not exceed 15.

题目大意 #

假设有从 1 到 N 的 N 个整数,如果从这 N 个数字中成功构造出一个数组,使得数组的第 i 位 (1 <= i <= N) 满足如下两个条件中的一个,我们就称这个数组为一个优美的排列。条件:

  • 第 i 位的数字能被 i 整除
  • i 能被第 i 位上的数字整除

现在给定一个整数 N,请问可以构造多少个优美的排列?

解题思路 #

  • 这一题是第 46 题的加强版。由于这一题给出的数组里面的数字都是不重复的,所以可以当做第 46 题来做。
  • 这题比第 46 题多的一个条件是,要求数字可以被它对应的下标 + 1 整除,或者下标 + 1 可以整除下标对应的这个数字。在 DFS 回溯过程中加入这个剪枝条件就可以了。
  • 当前做法时间复杂度不是最优的,大概只有 33.3%

代码 #


package leetcode

// 解法一 暴力打表法
func countArrangement1(N int) int {
	res := []int{0, 1, 2, 3, 8, 10, 36, 41, 132, 250, 700, 750, 4010, 4237, 10680, 24679, 87328, 90478, 435812}
	return res[N]
}

// 解法二 DFS 回溯
func countArrangement(N int) int {
	if N == 0 {
		return 0
	}
	nums, used, p, res := make([]int, N), make([]bool, N), []int{}, [][]int{}
	for i := range nums {
		nums[i] = i + 1
	}
	generatePermutation526(nums, 0, p, &res, &used)
	return len(res)
}

func generatePermutation526(nums []int, index int, p []int, res *[][]int, used *[]bool) {
	if index == len(nums) {
		temp := make([]int, len(p))
		copy(temp, p)
		*res = append(*res, temp)
		return
	}
	for i := 0; i < len(nums); i++ {
		if !(*used)[i] {
			if !(checkDivisible(nums[i], len(p)+1) || checkDivisible(len(p)+1, nums[i])) { // 关键的剪枝条件
				continue
			}
			(*used)[i] = true
			p = append(p, nums[i])
			generatePermutation526(nums, index+1, p, res, used)
			p = p[:len(p)-1]
			(*used)[i] = false
		}
	}
	return
}

func checkDivisible(num, d int) bool {
	tmp := num / d
	if int(tmp)*int(d) == num {
		return true
	}
	return false
}


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