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  • [leetcode]70. Climbing Stairs爬楼梯

    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.

    Input: 2
    Output: 2
    Explanation: There are two ways to climb to the top.
    1. 1 step + 1 step
    2. 2 steps
    -----------------------------------------------------------
    Input: 3
    Output: 3
    Explanation: There are three ways to climb to the top.
    1. 1 step + 1 step + 1 step
    2. 1 step + 2 steps
    3. 2 steps + 1 step

    题意:

    有n阶台阶,可以每次爬1步或者2步, 爬到终点有多少种不同的方法

    思路:

    一维dp

    用一个数组存下当前 k (k<=n)阶台阶有多少种不同的走法

    发现规律

    爬上n阶台阶 = 爬上n-1阶台阶再爬1阶 + 爬上n-2阶台阶再爬2阶

    爬上n-1阶台阶 =  爬上n-2阶台阶再爬1阶  +  爬上n-3阶台阶再爬2阶

    爬上n-2阶台阶 =  爬上n-3阶台阶再爬1阶  +  爬上n-4阶台阶再爬2阶

    初始化,

    dp[0] = 1

    dp[1] = 1

    转移方程,

    dp[i] = dp[i-1] + dp[i-2] 

    代码:

     1 class Solution {
     2     public int climbStairs(int n) {
     3         int [] dp = new int[n + 1];
     4         dp[0] = 1;
     5         dp[1] = 1;
     6         for(int i = 2; i <= n; i++){
     7             dp[i] = dp[i-1] + dp[i-2];
     8         }
     9         
    10         return dp[n];   
    11     }
    12 }
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  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9059494.html
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