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?
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
class Solution {
public int climbStairs(int n) {
//输入的是n个台阶,一次要么一步,要么两步,问有多少不同的步数,那就建立一个维护当前不同步数的数组
if(n <= 0)
return 0;
int[]dp = new int[n];
if(n == 1)
return 1;
else if(n == 2)
return 2;
dp[1] = 1;
dp[2] = 2;
for(int i = 3; i < n;i++){
dp[i] = dp[i-1]+dp[i-2];
}
return dp[n];
}
}