Algs4-1.4.12找出两个有序数组的公共元素-方法1

1.4.12编写一个程序,有序打印给定的两个有序数组(含有N个int值)中的所有公共元素,程序在最坏情况下所需的运行时间应该和N成比。
答:
图片
import java.util.Arrays;
public class TheSameElement
{
  public static void main(String[] args)
  {
     int[] a1=In.readInts(args[0]);
     int[] a2=In.readInts(args[1]);
     Arrays.sort(a1);
     Arrays.sort(a2);
     int a1lo=0;
     int a2lo=0;
     int a2KeyIndex=0;
     while (a1lo<a1.length && a2lo<a2.length)
     {
       a1lo=rankMax(a1,a1lo,a1.length-1,a1[a1lo]);
       a2KeyIndex=rankMax(a2,a2lo,a2.length-1,a1[a1lo]);
       if(a2KeyIndex!=-1)
       {
         StdOut.println(a1[a1lo]);
         a2lo=a2KeyIndex+1;
       }
       a1lo++;
     }
    
  }
 
   private static int rankMax(int[] a,int lo,int hi,int key)
  {
    int keyIndex=-1;
    while(lo<=hi)
    {
      int mid=lo+(hi-lo)/2;
      if (key<a[mid]) hi=mid-1;
      else if(key>a[mid]) lo=hi+1;
      else
      {
        keyIndex=mid;
        lo=mid+1;
      }//end if
    }//end while
      return keyIndex;
}//end rankMax
}

原文地址:https://www.cnblogs.com/longjin2018/p/9854418.html