70. Climbing Stairs
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?
经典的爬楼梯问题,共有n个台阶,每次只能向上迈一步或两步,问有多少种爬法。
反过来思考,如果站到最高处往下面看,要往下走,只有两种走法:迈一小步,迈两小步。假设从底层到达n-1台阶上有f(n-1)种走法,到达n-2台阶上有f(n-2)种走法,则到达n台阶的走法共有f(n)=f(n-1)+f(n-2).可用从高到低的递归方法也可以从底到高的递推方法。
相关阅读:漫画:什么是动态规划?里面第一个例子就是爬楼梯,讲解非常生动,图文并茂,把爬楼梯各种算法都分析的很有意思
class Solution_70_1 { public: int climbStairs(int n) { int result; int *arr = new int[n+1]; memset(arr, 0, sizeof(arr)); arr[1] = 1; arr[2] = 2; for (int i = 3; i <= n; i++) { arr[i] = arr[i - 1] + arr[i - 2]; } result = arr[n]; delete[] arr; return result; } }; //改用vector更方便更清晰 class Solution_70_1_ { public: int climbStairs(int n) { vector<int>steps(n+1); steps[1] = 1; steps[2] = 2; for (int i = 3; i <= n; i++) { steps[i] = steps[i - 1] + steps[i - 2]; } return steps[n]; } }; class Solution_70_2 { public: int climbStairs(int n) { int prev = 0; int cur = 1; for (int i = 1; i <= n; ++i) { int tmp = cur; cur += prev; prev = tmp; } return cur; } }; //递归方法 时间超时 提交失败 class Solution_70_3 { public: int climbStairs(int n) { return steps(n); } private: int steps(int n) { if (n == 1){ return 1; } else if (n == 2) { return 2; } else { return steps(n - 1) + steps(n - 2); } } };
//提交后beats率才2.21%,搞不懂。。。 class Solution_70_4 { public: int climbStairs(int n) { const double s = sqrt(5); //标准库函数floor,返回不大于给定变量的最大整数,截断 return floor((pow((1 + s) / 2, n + 1) + pow((1 - s) / 2, n + 1)) / s + 0.5); } };
最后一种方法用公式直接算: