Climbing Stairs：步长1或2到达终点

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2Output:  2Explanation:  There are two ways to climb to the top.1. 1 step + 1 step2. 2 steps

Example 2:

Input: 3Output:  3Explanation:  There are three ways to climb to the top.1. 1 step + 1 step + 1 step2. 1 step + 2 steps3. 2 steps + 1 step

思路：定眼一看，动态规划或者组合数，但组合数没发现什么直接规律，于是手写了几个，如：

1 -> 1

2 -> 2

3 -> 3

4 -> 5

5 -> 8

6 -> 13

菲波那切数列。

所以：

class Solution {    public int climbStairs(int n) {        int a = 1;        int b = 2;        int c = 0;        if(n == 1) return a;        if(n == 2) return b;        if(n == 0) return 0;        for(int i = 3 ;i <= n;i++){            c = a + b;            a = b;            b = c;        }        return c;    }}

既然是菲波那切数列，也可以矩阵快速幂进行运算

可以参见之前的博客：http://blog.csdn.net/u013300579/article/details/78027360无力吐槽排版。。。

public class Solution {    public int climbStairs(int n) {        int[][] q = {{1, 1}, {1, 0}};        int[][] res = pow(q, n);        return res[0][0];    }    public int[][] pow(int[][] a, int n) {        int[][] ret = {{1, 0}, {0, 1}};        while (n > 0) {            if ((n & 1) == 1) {                ret = multiply(ret, a);            }            n >>= 1;            a = multiply(a, a);        }        return ret;    }    public int[][] multiply(int[][] a, int[][] b) {        int[][] c = new int[2][2];        for (int i = 0; i < 2; i++) {            for (int j = 0; j < 2; j++) {                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];            }        }        return c;    }}

当然还有公式

F​n​​=1/√​5​​​[(​2​​1+√​5​​​​​)​n​​−(​2​​1−√​5​​​​​)​n​​]

public class Solution {    public int climbStairs(int n) {        double sqrt5=Math.sqrt(5);        double fibn=Math.pow((1+sqrt5)/2,n+1)-Math.pow((1-sqrt5)/2,n+1);        return (int)(fibn/sqrt5);    }}