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.
Example 1:
Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
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
题目大意:
给出n个阶梯,每次只能上1或者2级台阶,输出存在多少组组合。
解题思路:
应该说是规律题,模拟一下前5个答案,就不难发现n的答案是n-1和n-2答案的和。
从题目的角度看,可以这样解释我们知道走到第n级台阶,只有两种方法从n-1级台阶走一步,或者从n-2级台阶走两步。所以其存在的组合即为n-1和n-2答案的和。
class Solution {
public:
int climbStairs(int n) {
if(n==0||n==1||n==2){
return n;
}
int l = 1;
int r = 2;
int ans;
for(int i=3;i<=n;i++){
ans=r+l;
l = r;
r = ans;
}
return ans;
}
};