LeetCode: Climbing Stairs [070]
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【题目】
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?
【题意】
有个梯子有n阶,每次只能爬1阶或者2阶,为爬到梯子顶共有多少种爬法
【思路】
依次确定跳到每一阶上的爬法数目
这其实是一个斐波那契数列数列。假设A[i]表示爬到第i阶的爬法数
当i=1时,A[1]=1 (从梯子底只要爬一阶即可)
当i=2时,A[2]=2 (从第一阶往上再爬一阶,或者从梯子底直接爬谅解,所以是两种爬法)
当i>=3时,A[i]=A[i-1]+A[i-2] (因为每次只能爬一阶或者两阶,因此要跳到当前阶,只可能从他的前一阶或者前前一阶往上爬)
【代码】
class Solution {public: int climbStairs(int n) { if(n==0)return 0; if(n==1)return 1; if(n==2)return 2; vector<int>stairs(n, 0); stairs[0]=1; stairs[1]=2; for(int i=2; i<n; i++){ stairs[i]=stairs[i-1]+stairs[i-2]; } return stairs[n-1]; }};
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