LeetCode-70-Climbing Stairs(爬楼梯)

Q:

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.

Analysis:

经典的动态规划问题。一阶台阶只可能有1种方法，即只上一层台阶；二层台阶可以有2种方法，即一次上一阶台阶上两次或者一次上两阶台阶；依次类推。

公式为:dp[i]=dp[i-1]+dp[i-2]

Code:

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