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);
        }
    }
}