【动态规划】Leetcode编程题解:70. Climbing Stairs
2017-05-07 16:36
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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?
提示:Given n will be a positive integer.
爬到顶层只和第n - 1阶和第n - 2阶有关,刚好符合斐波那契数列的特性,所以可以考虑用斐波那契数列解决。
解法1:
但是这种递归的方法在n > 43时候会超时,所以适用范围比较小。
解法2:
将第n - 1阶和第n阶用数组进行存储,可以极大缩短计算时间。
解法3:
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?
提示:Given n will be a positive integer.
爬到顶层只和第n - 1阶和第n - 2阶有关,刚好符合斐波那契数列的特性,所以可以考虑用斐波那契数列解决。
解法1:
class Solution { public: int climbStairs(int n) { if(n == 1) return 1; if(n == 2) return 2; if(n > 2) return climbStairs(n - 2) + climbStairs(n - 1); } };
但是这种递归的方法在n > 43时候会超时,所以适用范围比较小。
解法2:
class Solution { public: int climbStairs(int n) { int ways[2] = {1, 1}; for(int i = 1; i < n; i++) { int temp = ways[1]; ways[1] += ways[0]; ways[0] = temp; } return ways[1]; } };
将第n - 1阶和第n阶用数组进行存储,可以极大缩短计算时间。
解法3:
class Solution { public: int climbStairs(int n) { n++; double root5 = pow(5, 0.5); double result = 1/root5*( pow((1+root5)/2, n) - pow((1-root5)/2, n) ); return (int)(result); } };直接套用斐波那契数列的通项公式。
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