[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.

 

public class Solution {
    public int climbStairs2(int n) {
        int[] f = new int[n + 1];
        f[0] = 1;
        
        for(int i = 1; i <= n; i++){
            if(i - 3 >= 0){
                f[i] = f[i - 1] + f[i - 2] + f[i - 3];
            }
            else if(i - 2 >= 0){
                f[i] = f[i - 1] + f[i - 2];
            }
            else if(i - 1 >= 0){
                f[i] = f[i - 1];
            }
        }
        return f[n];
    }
}

 

Related Problems

Climbing Stairs