1、public static String exR1(int n){ if(n<=0) return ""; return exR1(n-3)+ n + exR1(n-2) + n; } System.out.println(exR1(6)); //311361142246
2、public static String exR2(int n){ String s = exR1(n-3)+ n + exR1(n-2) + n; if(n<=0) return ""; return s; } ··注: 这段代码的基础情况永远不会被访问,会循环往复直到发生 StackOverflowError.
3、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;
}
时间: 2024-11-10 01:04:38