70. Climbing Stairs

原创于 2017-10-29 18:53:00 发布 · 293 阅读

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本文探讨了经典的爬楼梯问题，即如何用不同方式爬到楼梯顶部的问题。使用动态规划的方法，提供了Java和Python两种语言的实现方案，通过递推公式计算达到第n级台阶的不同方式的数量。

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.

java

class Solution {
    public int climbStairs(int n) {
        if (n == 0) {
            return 0;
        }
        if (n == 1) {
            return 1;
        }
        if (n == 2) {
            return 2;
        }
        int[] f = new int[n];
        f[0] = 1;
        f[1] = 2;
        for (int i = 2; i < n; i++) {
            f[i] = f[i - 1] + f[i - 2];
        }
        return f[n - 1];
    }
}

python

class Solution:
    """
    @param n: An integer
    @return: An integer
    """
    def climbStairs(self, n):
        # write your code here
        if n == 0:
            return 0
        if n == 1:
            return 1
        if n == 2:
            return 2
        res = [1, 2]
        for i in range(2, n):
            res.append(res[i - 1] + res[i - 2])
        return res[len(res) - 1]