[LintCode] Climbing Stairs

最新推荐文章于 2020-03-10 04:29:12 发布

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最新推荐文章于 2020-03-10 04:29:12 发布

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原文链接：https://segmentfault.com/a/1190000004953759

本文探讨了经典的爬楼梯问题，给出了两种解决方案：一种基于斐波那契数列的迭代算法，另一种则是动态规划的方法。这两种方法都有效地解决了如何计算达到楼梯顶部的不同方式的数量。

Problem

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?

Example

Given an example n=3 , 1+1+1=2+1=1+2=3

return 3

Note

无需动规，无需额外空间，等同于菲波那切数列。
当然噜，也可以动规，记住dp[0] = 1就好。

Solution

Fibonacci

public class Solution {
    public int climbStairs(int n) {
        if (n < 2) return 1;
        int one = 1, two = 1, total = 0;
        for (int i = 2; i <= n; i++) {
            res = one + two;
            two = one;
            one = res;
        }
        return res;
    }
}

DP

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