Ex 2_22 两个有序列表合并后的第k小元素..._第四次作业

 package org.xiu68.ch02;

 public class Ex2_22 {

     public static void main(String[] args) {
         // TODO Auto-generated method stub
         //两数组有序，寻找两数组合并后第k小元素，O(logm+logn)
         int[] a=new int[]{1,3,5,7,9,11,13,15,17,19};
         int[] b=new int[]{0,2,4,6,8,10,12,14,16,18};

         for(int k=1;k<11;k++){
             int result=find(a, b, 0, a.length-1, 0, b.length-1, k);
             System.out.println(result);
         }
     }

     public static int find(int a[],int b[],int aStart,int aEnd,int bStart,int bEnd,int k){
         int aMid = (aStart + aEnd) / 2;    //a数组中间元素位置
         int bMid = (bStart + bEnd) / 2;    //b数组中间元素位置
         if (aStart > aEnd)                //a数组要计算元素个数为0
             return b[bStart+k-1];
         if (bStart > bEnd)                //b数组要计算元素个数为0
             return a[aStart+k-1];    

         if (a[aMid] <= b[bMid]) {        //b[中间位置]大于a[中间位置]的情况
             if (k <= (aMid-aStart) + (bMid-bStart) + 1)    //k小于两数组合并后的中间位置
                 return find(a,b,aStart, aEnd, bStart, bMid-1, k);
             else                                        //k大于两数组合并后的中间位置
                 return find(a,b,aMid+1, aEnd, bStart, bEnd, k-(aMid-aStart)-1);
         } else {                        //a[中间位置]大于b[中间位置]的情况
             if (k <= (aMid-aStart) + (bMid-bStart) + 1)
                 return find(a,b,aStart, aMid-1, bStart, bEnd, k);
             else
                 return find(a,b,aStart, aEnd, bMid+1, bEnd, k-(bMid-bStart)-1);
         }
     }
 }