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];
}
}
}