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?

?

方法一：遞歸

public int climbStairs2(int n) {if (n == 0)return 1;if (n < 3)return n;return climbStairs(n - 1) + climbStairs(n - 2);}

?

方法二：用臨時(shí)變量保存前面兩個(gè)數(shù)

public int climbStairs(int n) {if (n <= 0)return n == 0 ? 1 : 0;int result = 0;if (n < 3)return n;int pre = 1;int last = 2;for (int i = 3; i <= n; i++) {result = last + pre;pre = last;last = result;}return result;}

總結(jié)：方法一之所以能夠優(yōu)化為方法二是因?yàn)榉椒ㄒ挥?jì)算了f(n-1)和f(n-2)（注意：這里f（n）指的是爬上第n階的方法數(shù))，因此我們用兩個(gè)臨時(shí)變量保存就好了

與這題類(lèi)似的另外一個(gè)題是：

There is a fence with?n?posts, each post can be painted with one of thek?colors.
You have to paint all the posts such that no more than two adjacent fence posts have the same color.
Return the total number of ways you can paint the fence.

該題也是類(lèi)似的原理

public class Solution {/*** @param n non-negative integer, n posts* @param k non-negative integer, k colors* @return an integer, the total number of ways*/public int numWays(int n, int k) {if (n == 0 || k == 0)return 0;if (n < 3) {return n == 1 ? k : k * k;}int pre = k;int last = k * k;int result = 0;for (int i = 3; i <= n; n++) {result = (k - 1) * pre + (k - 1) * last;pre = last;last = result;}return result;} }

?

轉(zhuǎn)載于:https://www.cnblogs.com/googlemeoften/p/5833911.html

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