[LeetCode]70.Climbing Stairs

【题目】

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?

【题意】

爬楼梯。爬到楼顶需要n步。

每一次你都可以爬一步或者爬两步，问有多少种方式爬到楼顶？

【分析】

设 f (n) 表示爬 n 阶楼梯的不同方法数，为了爬到第 n 阶楼梯，有两个选择：
• 从第 n - 1 阶前进 1 步；
• 从第 n - 2 阶前进 2 步；
因此，有 f (n) = f (n - 1) + f (n - 2)。

这是一个斐波那契数列。

详细分析请参考：编程之美之斐波那契数列

【代码1】

// 递归
class Solution {
public:int climbStairs(int n) {return Fibonacci(n);}
private:int Fibonacci(int n){if(n <= 2){return n;}return Fibonacci(n - 1) + Fibonacci(n - 2);}
};

【代码2】

/*********************************
*   日期：2014-01-23
*   作者：SJF0115
*   题号: Climbing Stairs
*   来源：http://oj.leetcode.com/problems/climbing-stairs/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <stdio.h>
#include <vector>
using namespace std;
// 迭代，时间复杂度 O(n)，空间复杂度 O(1)
class Solution {
public:int climbStairs(int n) {int prev = 0;int cur = 1;for(int i = 1; i <= n ; ++i){int tmp = cur;cur = prev + cur;prev = tmp;}return cur;}
};
int main() {Solution solution;int result;result = solution.climbStairs(40);printf("Result:%d\n",result);return 0;
}

【代码3】

/*********************************
*   日期：2014-01-23
*   作者：SJF0115
*   题号: Climbing Stairs
*   来源：http://oj.leetcode.com/problems/climbing-stairs/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
// 数学公式，时间复杂度 O(1)，空间复杂度 O(1)
class Solution {
public:int climbStairs(int n) {double s = sqrt(5);return floor((pow((1+s)/2, n+1) + pow((1-s)/2, n+1))/s + 0.5);}
};
int main() {Solution solution;int result;result = solution.climbStairs(40);printf("Result:%d\n",result);return 0;
}

【代码4】

/*********************************
*   日期：2014-01-24
*   作者：SJF0115
*   题号: Climbing Stairs
*   来源：http://oj.leetcode.com/problems/climbing-stairs/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <stdio.h>
using namespace std;class Solution {
public://Fibonacci数列int climbStairs(int n) {if(n == 0){return 0;}int A[2][2] = {1,1,1,0};//初始为单位矩阵int Matrix[2][2] = {1,0,1,0};//n = n - 1;for(; n ;n >>= 1){//奇偶if(n&1){MatrixMulti(Matrix,A);}//ifMatrixMulti(A,A);}//forreturn Matrix[0][0];}
private://矩阵乘法void MatrixMulti(int matrix[2][2],int matrix2[2][2]){int a = matrix[0][0] * matrix2[0][0] +  matrix[0][1] * matrix2[1][0];int b = matrix[0][0] * matrix2[0][1] +  matrix[0][1] * matrix2[1][1];int c = matrix[1][0] * matrix2[0][0] +  matrix[1][1] * matrix2[1][0];int d = matrix[1][0] * matrix2[0][1] +  matrix[1][1] * matrix2[1][1];matrix[0][0] = a;matrix[0][1] = b;matrix[1][0] = c;matrix[1][1] = d;}
};int main() {Solution solution;int result;for(int i = 1;i < 10;i++){result = solution.climbStairs(i);printf("Result:%d\n",result);}return 0;
}

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