LeetCode——Climbing Stairs
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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?
原题链接:https://oj.leetcode.com/problems/climbing-stairs/
题目:你在爬楼梯。需要 n 步才能到顶部。
每次你爬1 或 2 步。有多少种独立的爬到顶部的方式?
思路:首先很容易就想到了递归的解法。但是超时了。
public int climbStairs(int n) {if(n < 0)return 0;if(n <= 1)return 1;return climbStairs(n - 1) + climbStairs(n - 2);}
所以采用非递归的方式,其实此题类似于求斐波那契数列的和,但是递归不仅慢还可能溢出。下面采用非递归的方法,其中pre代表前n-1台阶的方法数,current代表第n台阶的方法数。
public int climbStairs(int n) {if (n == 0 || n == 1)return 1;int pre = 1;int current = 1;for (int i = 2; i <= n; i++) {int temp = current + pre;pre = current;current = temp;}return current;}
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