[LintCode] Climbing Stairs II

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

Almost identical with Climbing Stairs.

 1 public class Solution {
 2     public int climbStairs2(int n) {
 3         int[] f = new int[n + 1];
 4         f[0] = 1;
 5         
 6         for(int i = 1; i <= n; i++){
 7             if(i - 3 >= 0){
 8                 f[i] = f[i - 1] + f[i - 2] + f[i - 3];
 9             }
10             else if(i - 2 >= 0){
11                 f[i] = f[i - 1] + f[i - 2];
12             }
13             else if(i - 1 >= 0){
14                 f[i] = f[i - 1];
15             }
16         }
17         return f[n];
18     }
19 }