59 lines
1.3 KiB
Java
59 lines
1.3 KiB
Java
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# 10.4 变态跳台阶
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[NowCoder](https://www.nowcoder.com/practice/22243d016f6b47f2a6928b4313c85387?tpId=13&tqId=11162&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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一只青蛙一次可以跳上 1 级台阶，也可以跳上 2 级... 它也可以跳上 n 级。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。
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<img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/cd411a94-3786-4c94-9e08-f28320e010d5.png" width="380px">
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## 解题思路
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### 动态规划
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```java
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public int JumpFloorII(int target) {
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int[] dp = new int[target];
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Arrays.fill(dp, 1);
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for (int i = 1; i < target; i++)
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for (int j = 0; j < i; j++)
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dp[i] += dp[j];
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return dp[target - 1];
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}
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```
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### 数学推导
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跳上 n-1 级台阶，可以从 n-2 级跳 1 级上去，也可以从 n-3 级跳 2 级上去...，那么
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```
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f(n-1) = f(n-2) + f(n-3) + ... + f(0)
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```
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同样，跳上 n 级台阶，可以从 n-1 级跳 1 级上去，也可以从 n-2 级跳 2 级上去... ，那么
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```
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f(n) = f(n-1) + f(n-2) + ... + f(0)
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```
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综上可得
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```
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f(n) - f(n-1) = f(n-1)
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```
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即
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```
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f(n) = 2*f(n-1)
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```
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所以 f(n) 是一个等比数列
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```source-java
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public int JumpFloorII(int target) {
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return (int) Math.pow(2, target - 1);
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}
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```
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