[lintcode easy]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?

 

Example

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

return 3

 

 

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分析：有四个台阶的时候，就是三个台阶的上法+1步，和2个台阶的上法+2步。

因此f(n)=f(n-1)+f(n-2);

 

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

 

转载于:https://www.cnblogs.com/kittyamin/p/5011852.html