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?
经典动态规划的问题，如果到最后一个的话，有从倒数第一节和倒数第二节，两种走到的放法。
public class Solution{ public int climbStairs(int n){ if(n == 0 || n == 1 || n ==2){ return n; } int[] steps = new int[n+1]; steps[0] = 0; steps[1] = 1； steps[2] = 2; for(int i = 3; i <= n ; i++){ steps[i] = steps[i-1] + steps[i-2]; } return steps[n]; } }
转载于:https://www.cnblogs.com/hewx/p/4532427.html
