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.
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
题意：
有n阶台阶，可以每次爬1步或者2步， 爬到终点有多少种不同的方法
思路：
一维dp
用一个数组存下当前 k (k<=n)阶台阶有多少种不同的走法
发现规律
爬上n阶台阶 = 爬上n-1阶台阶再爬1阶 + 爬上n-2阶台阶再爬2阶
爬上n-1阶台阶 = 爬上n-2阶台阶再爬1阶 + 爬上n-3阶台阶再爬2阶
爬上n-2阶台阶 = 爬上n-3阶台阶再爬1阶 + 爬上n-4阶台阶再爬2阶
初始化，
dp[0] = 1
dp[1] = 1
转移方程，
dp[i] = dp[i-1] + dp[i-2]
代码：
1 class Solution { 2 public int climbStairs(int n) { 3 int [] dp = new int[n + 1]; 4 dp[0] = 1; 5 dp[1] = 1; 6 for(int i = 2; i <= n; i++){ 7 dp[i] = dp[i-1] + dp[i-2]; 8 } 9 10 return dp[n]; 11 } 12 }
转载于:https://www.cnblogs.com/liuliu5151/p/9059494.html
