标签:斐波那契 tput positive turn mat 没有 tair tin integer
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.
Example 1:
Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
Input: 3 Output: 3 Explanation: There are three ways to climb to the tNop. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step
这个题很显然是用动态规划的做法,人在跳最后一步时,有两种可能,人可能跳一个台阶,也可能跳两个台阶,如果说n个台阶的走法是F(n),那么F(n) = F(n-1) + F(n-2),当n=1时,F(n) = 1,当n = 2时,F(n) = 2;
public int climbStairs1(int n) { if (n <= 0) return -1; else if (n == 1) return 1; else if (n == 2) return 2; else return climbStairs1(n - 1) + climbStairs1(n - 2); }
这种解法时间复杂度很糟糕,在编译器中,每次递归计算的值都没有被保存下来,不建议采用。为了每次递归计算的值保存下来,我们创建一个空间大小为n的数组dp[],空间和时间复杂度都是O(n)。
public int climbStairs2(int n) { if (n <= 0) return -1; else if (n == 1) return 1; int[] dp = new int[n + 1]; dp[1] = 1; dp[2] = 2; for (int i = 3; i <= n; i++) dp[i] = dp[i - 1] + dp[i - 2]; return dp[n]; }
在计算的过程中我们发现,dp[]数组满足斐波那契数列关系,可以利用这一特点只定义三个变量,由此来节省空间复杂度,使得空间复杂度优化到O(n).
public int climbStairs3(int n) { if (n == 1) return 1; int first = 1; int second = 2; for (int i = 3; i <= n; i++) { int third = first + second; first = second; second = third; } return second; }
标签:斐波那契 tput positive turn mat 没有 tair tin integer
原文地址:https://www.cnblogs.com/tiandiou/p/9692802.html