2019-11-02 12:07:41 +08:00
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# 10.3 跳台阶
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2019-11-03 23:57:08 +08:00
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## 题目链接
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2019-11-02 12:07:41 +08:00
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[NowCoder](https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4?tpId=13&tqId=11161&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。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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2019-12-05 01:47:57 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/9dae7475-934f-42e5-b3b3-12724337170a.png" width="380px"> </div><br>
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2019-11-02 12:07:41 +08:00
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## 解题思路
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当 n = 1 时,只有一种跳法:
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2019-12-05 01:47:57 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/72aac98a-d5df-4bfa-a71a-4bb16a87474c.png" width="250px"> </div><br>
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2019-11-02 12:07:41 +08:00
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当 n = 2 时,有两种跳法:
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2019-12-05 01:47:57 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/1b80288d-1b35-4cd3-aa17-7e27ab9a2389.png" width="300px"> </div><br>
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2019-11-02 12:07:41 +08:00
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跳 n 阶台阶,可以先跳 1 阶台阶,再跳 n-1 阶台阶;或者先跳 2 阶台阶,再跳 n-2 阶台阶。而 n-1 和 n-2 阶台阶的跳法可以看成子问题,该问题的递推公式为:
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2019-12-05 01:47:57 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/508c6e52-9f93-44ed-b6b9-e69050e14807.jpg" width="350px"> </div><br>
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2019-11-02 12:07:41 +08:00
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```java
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public int JumpFloor(int n) {
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if (n <= 2)
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return n;
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int pre2 = 1, pre1 = 2;
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int result = 0;
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for (int i = 2; i < n; i++) {
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result = pre2 + pre1;
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pre2 = pre1;
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pre1 = result;
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}
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return result;
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}
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```
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2019-11-02 14:39:13 +08:00
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2019-11-02 17:33:10 +08:00
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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