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Recursion.java
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/
Recursion.java
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package com.kyson.chapter1.section1;
import edu.princeton.cs.algs4.StdOut;
/***
*
*
*
*/
public class Recursion {
/**
* 1.1.16 给出 exR1(6) 的返回值:
*/
public static String exR1(int n) {
if (n <= 0)
return "";
return exR1(n - 3) + n + exR1(n - 2) + n;
}
/***
* 1.1.18 请看以下递归函数: mystery(2, 25) 和 mystery(3, 11) 的返回值是多少?给定正整数 a 和
* b，mystery(a,b)计算的结果是什么?将代码中的 + 替换为 * 并将 return 0 改为 return 1，然后回答相同 的问题。
*/
public static int mystery(int a, int b) {
if (b == 0)
return 0;
if (b % 2 == 0)
return mystery(a + a, b / 2);
return mystery(a + a, b / 2) + a;
}
public static int mystery1(int a, int b) {
if (b == 0)
return 1;
if (b % 2 == 0)
return mystery1(a * a, b / 2);
return mystery1(a * a, b / 2) * a;
}
public static void main(String[] args) {
/***
* 1.1.16
*/
System.out.println("1.1.16:");
System.out.println("exR1(6)的输出：" + exR1(6));
/***
* 1.1.18
*/
System.out.println("1.1.18:");
System.out.println("mystery(2, 25):" + mystery(2, 25)); // 输出50
System.out.println("mystery(3, 11):" + mystery(3, 11)); // 输出33
System.out.println("mystery1(2, 25):" + mystery1(2, 25)); // 输出33554432
System.out.println("mystery1(3, 11):" + mystery1(3, 11)); // 输出177147
/***
* 1.1.19
*/
System.out.println("1.1.19");
for (int N = 0; N < 100; N++)
StdOut.println(N + " " + Fibonacci.F1(N));
}
}