Saturday, August 22, 2020

526. Beautiful Arrangement -----------M

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 ith position is divisible by i.
  2. i is divisible by the number at the ith position.

 

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.


A:

Idea is straightforward,  just swap and check position.

class Solution {
public:
    int countArrangement(int N) {
        vector<int> nums(N+1,0);
        for(int i=1;i<=N;i++ ){
            nums[i]=i;
        }
        int res = 0;
        helper(nums, res, 1);
        return res;
    }
private:
    void helper(vector<int> & nums, int &res,  int start){
        if(start >= nums.size()){
            res++;
            return;
        }
        for(int end = start; end< nums.size();end++){
            if( nums[end] % start ==0 || start % nums[end]==0){
                // swap end with start
                if(end != start){
                    int tmp = nums[start];
                    nums[start] = nums[end];
                    nums[end] = tmp;
                }
                helper(nums, res, start+1);
                // swap back
                if(end != start){
                    int tmp = nums[start];
                    nums[start] = nums[end];
                    nums[end] = tmp;
                }
            }
        }
    }
};


犯了一个极其SB的错误,  swap的时候, nums[end] = tmp 写成 nums[end] = start;

就是为了图快,哎。

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