[算法专栏] 爬楼梯问题

int Fibonacci1(unsigned int N)  

{  

    if(N<=2)  

        return N;  

    int fibtwo=2;  

    int fibone=1;  

    int fibN=0;  

    for(unsigned int i=3;i<=N;i++)  

    {  

        fibN=fibone+fibtwo;  

        fibone=fibtwo;  

        fibtwo=fibN;  

    }  

    return fibN;  

}  

  1 /**
  2  * 爬楼梯问题: 实质就是斐波那契数列.
  3  * 
  4  * @author X-knight
  5  *
  6  */
  7 public class ClimbTheStairs {
  8     int total;
  9 
 10     // 递归调用
 11     public int fib01(int n) {
 12         if (n == 1 || n == 2)
 13             total = n;
 14         else
 15             total = fib01(n - 2) + fib01(n - 1);
 16         return total;
 17     }
 18 
 19     // 三目运算符
 20     public int fib02(int n) {
 21         return (n == 1 || n == 2) ? n : fib02(n - 2) + fib02(n - 1);
 22     }
 23 
 24     // 备忘录法
 25     public int dfs(int n, int[] array) {
 26         if (array[n] != 0) {
 27             return array[n];
 28         } else {
 29             array[n] = dfs(n - 1, array) + dfs(n - 2, array);
 30             return array[n];
 31         }
 32     }
 33 
 34     public int fib03(int n) {
 35         if (n == 0) {
 36             return 1;
 37         }
 38         if (n == 1 || n == 2) {
 39             return n;
 40         } else {
 41             int[] array = new int[n + 1];
 42             array[1] = 1;
 43             array[2] = 2;
 44             return dfs(n, array);
 45         }
 46     }
 47 
 48     // 动态规划法 (利用数组来存储)
 49     public int fib04(int n) {
 50         if (n == 0)
 51             return 1;
 52         int[] array = new int[n + 1];
 53         array[0] = 1;
 54         array[1] = 1;
 55         for (int i = 2; i <= n; i++) {
 56             array[i] = array[i - 1] + array[i - 2];
 57         }
 58         return array[n];
 59     }
 60 
 61     // 状态压缩法(又称滚动数组、滑动窗口，用于优化动态规划法的空间复杂度)
 62     public int fib05(int n) {
 63         if (n == 1 || n == 0)
 64             return 1;
 65         n = n - 1;
 66         int result = 0;
 67         int zero = 1;
 68         int first = 1;
 69         while (n > 0) {
 70             result = zero + first;
 71             zero = first;
 72             first = result;
 73             n--;
 74         }
 75         return result;
 76     }
 77 
 78     // 斐波那契数列的通项公式
 79     public int fib06(int n) {
 80         if (n == 0)
 81             return 1;
 82         if (n == 1 || n == 2)
 83             return n;
 84         int result = (int) Math.floor(
 85                 1 / Math.sqrt(5) * (Math.pow((1 + Math.sqrt(5)) / 2, n + 1) - Math.pow((1 - Math.sqrt(5)) / 2, n + 1)));
 86         return result;
 87     }
 88 
 89     @Test
 90     public void testClimb() {
 91         int num01 = fib01(7);
 92         System.out.println(num01);
 93         int num02 = fib02(8);
 94         System.out.println(num02);
 95         int num03 = fib03(0);
 96         System.out.println(num03);
 97         int num04 = fib04(8);
 98         System.out.println(num04);
 99         int num05 = fib05(8);
100         System.out.println(num05);
101         int num06 = fib06(2);
102         System.out.println(num06);
103     }
104 }