1. 题目描述
/** 阿克曼(Ackmann)函数 【题目描述】 阿克曼(Ackmann)函数A(m,n)中,m,n定义域是非负整数(m<=3,n<=10),函数值定义为: akm(m,n) = n+1; (m=0时) akm(m,n) = akm(m-1,1); (m>0,n=0时) akm(m,n) = akm(m-1,akm(m, n-1)); (m,n>0时) 【输入】输入m和n。 【输出】函数值。 【输入样例】2 3 【输出样例】9 */
写出递归与非递归算法,并输出调用过程。
2. 代码实现递归和非递归
import java.util.Stack; public class Main2 { public static void main( String[] args ) { System.out.println(AkmRecur(2, 3)); System.out.println(AkmNonRecur(2, 3)); } //递归 public static int AkmRecur(int m, int n) { if (m == 0) return n + 1; else if (n == 0) return AkmRecur(m - 1, 1); else return AkmRecur(m - 1, AkmRecur(m, n - 1)); } //非递归 public static int AkmNonRecur(int m, int n) { Stack<Integer> sx = new Stack<Integer>(); Stack<Integer> sy = new Stack<Integer>(); int x = 0; int y = 0; sx.push(m); sy.push(n); while ((!sx.empty()) && (!sy.empty())) { if (sx.peek() != 0 && sy.peek() == 0) { x = sx.peek(); y = sy.peek(); sx.pop(); sy.pop(); sx.push(x-1); sy.push(1); }else if (sx.peek() != 0 && sy.peek() != 0) { while (sx.peek()!= 0 &&sy.peek()!= 0) { x = sx.peek(); y = sy.peek()-1; sx.pop(); sy.pop(); sx.push(x-1); sy.push(-1); sx.push(x); sy.push(y); } } else { y = sy.peek(); sx.pop(); if (sx.empty()){ return y + 1; } else { sy.pop(); sy.pop(); sy.push(y+1); } } } return -1; } }