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3 files changed

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‎lcof/面试题62. 圆圈中最后剩下的数字/README.md‎

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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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设 `f(n, m)` 表示从 n 个数中每次删除第 m 个,最后剩下的数字。
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第一次删除第 m 个,剩下 `n-1` 个数,那么 `x = f(n - 1, m)` 就表示从 n-1 个数中每次删除第 m 个,最后剩下的数字。
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我们求得 x 之后,便可以知道,`f(n, m)` 应该是从 `m % n` 开始数的第 x 个元素,即 `f(n, m) = ((m % n) + x) % n`。
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当 n 为 1 时,最后留下的数字序号一定为 0。
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递归求解即可,也可以改成迭代。
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### Python3
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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递归版本:
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```python
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class Solution:
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def lastRemaining(self, n: int, m: int) -> int:
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def f(n, m):
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if n == 1:
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return 0
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x = f(n - 1, m)
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return (m + x) % n
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return f(n, m)
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```
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迭代版本:
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```python
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class Solution:
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def lastRemaining(self, n: int, m: int) -> int:
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f = 0
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for i in range(2, n + 1):
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f = (f + m) % i
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return f
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```
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### Java
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int lastRemaining(int n, int m) {
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int f = 0;
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for (int i = 2; i <= n; ++i) {
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f = (f + m) % i;
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}
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return f;
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}
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}
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```
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### ...
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class Solution {
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public int lastRemaining(int n, int m) {
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int f = 0;
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for (int i = 2; i <= n; ++i) {
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f = (f + m) % i;
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}
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return f;
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}
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}
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class Solution:
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def lastRemaining(self, n: int, m: int) -> int:
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f = 0
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for i in range(2, n + 1):
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f = (f + m) % i
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return f

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