资讯专栏INFORMATION COLUMN

[LintCode] Climbing Stairs II

chengtao1633 / 1650人阅读

Problem

A child is running up a staircase with n steps, and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs.

Example

n=3
1+1+1=2+1=1+2=3=3

return 4

Solution
public class Solution {
    /*
     * @param n: An integer
     * @return: An integer
     */
    public int climbStairs2(int n) {
        // write your code here
        int[] dp = new int[n+3];
        dp[0] = 1;
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i-1] + dp[i-2] + dp[i-3];
        }
        return dp[n];
    }
};

文章版权归作者所有,未经允许请勿转载,若此文章存在违规行为,您可以联系管理员删除。

转载请注明本文地址:https://www.ucloud.cn/yun/68158.html

相关文章

  • [LintCode] Climbing Stairs

    摘要: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,记住就好。 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 ...

    jemygraw 评论0 收藏0
  • [Leetcode] Climbing Stairs 爬楼梯

    摘要:递归法复杂度时间空间思路这题几乎就是求解斐波那契数列。最简单的方法就是递归。但重复计算时间复杂度高。代码递推法复杂度时间空间思路实际上我们求的时候只需要和的值,所以可以减少一些空间啊。 Climbing Stairs You are climbing a stair case. It takes n steps to reach to the top. Each time you c...

    tinyq 评论0 收藏0
  • [leetcode] Climbing Stairs

    摘要:实质就是把之前出现过的中间结果记录,下次再出现相同情况的时候,通过可以只用的时间复杂度得到。表示到达第层楼梯的不同走法。那么题目中每次可以选择走一步,或者两步,。从迭代公式可以知道,有两个,和。 70. Climbing Stairs You are climbing a staircase. It takes n steps to reach to the top.Each tim...

    int64 评论0 收藏0
  • 746. Min Cost Climbing Stairs

    746. Min Cost Climbing Stairs On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find mini...

    skinner 评论0 收藏0
  • 070. 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? Note: Given n will be a positive i...

    jay_tian 评论0 收藏0

发表评论

0条评论

最新活动
阅读需要支付1元查看
<