LeetCode Beautiful Arrangement
原题链接在这里:https://leetcode.com/problems/beautiful-arrangement/description/
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:
- The number at the ith position is divisible by i.
- 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:
- N is a positive integer and will not exceed 15.
题解:
典型的backtracking. 用visited记录用过的位置, 挨个position往后permute. 遇到不合规矩的直接返回, 看能不能走到N.
Time Complexity: exponential. 每次走到最后才遇到不和规矩的backtrack回来.
Space: O(N).
AC Java:
class Solution {
int res = 0;
public int countArrangement(int N) {
if(N <= 0){
return 0;
} boolean [] visited = new boolean[N+1];
dfs(visited, 1, N);
return res;
} private void dfs(boolean [] visited, int pos, int N){
if(pos > N){
res++;
return;
} for(int i = 1; i<=N; i++){
if(!visited[i] && (i%pos==0 || pos%i==0)){
visited[i] = true;
dfs(visited, pos+1, N);
visited[i] = false;
}
}
}
}
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