zoukankan      html  css  js  c++  java
  • leetcode 526. Beautiful Arrangement

    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.

    基本题:回溯法。

    解法一:开辟一个新的数组,标记是否访问过。

     1 class Solution {
     2     int cnt = 0;
     3 public:
     4     int countArrangement(int N) {
     5         vector<bool> visited(N + 1, false);
     6         calculate(N, visited, 1);
     7         return cnt;
     8     }
     9     void calculate(int N, vector<bool> &visited, int pos) {
    10         if (pos > N) {
    11             cnt++;
    12             return;
    13         }
    14         for (int i = 1; i <= N; ++i) {
    15             if (!visited[i] && ((pos % i == 0) || (i % pos == 0))) {
    16                 visited[i] = true;
    17                 calculate(N, visited, pos + 1);
    18                 visited[i] = false;
    19             } 
    20         }
    21     }
    22 };

    解法二:

  • 相关阅读:
    SSH-框架工作笔记
    Ajax基础
    Hibernate基础
    Struts1 中的国际化
    MyEclipse中的快捷键
    Oracle_存储过程
    oracle_开发子程序和包
    常用的正则表达式
    jQuery中的事件
    AcWing 1118. 分成互质组
  • 原文地址:https://www.cnblogs.com/qinduanyinghua/p/11904517.html
Copyright © 2011-2022 走看看