给定一个运算规则，让其在最短的步骤之内达到1

给定一个数字n,如果n是偶数就让n/2，如果是奇数就让n +1或-1，例
n = 31;
n = n - 1; //30
n = n / 2; //15
n = n + 1; //16
n = n / 2; //8
n = n / 2; //4
n = n / 2; //2
n = n / 2; //1
问：怎样能够以最少的计算次数，算出n = 1;

 

csdn一网友给的算法：

 

 

 1 public class Test1 {
 2 
 3     public static void main(String[] args) {
 4         int n = 31;
 5         
 6         if(n < 1) {
 7             return;
 8         }
 9         int count = 0;        
         while(n != 1) {            
             if(n % 2 == 0) {
                 n /= 2;                
             } else {
                 if((n & 2) == 0 || n == 3) {
                     n--;
                 } else {
                     n++;
                 }
             }
             count++;
             System.out.println(count + ": " + n);
         }
         System.out.println("Total: " + count);
     }
 }