Algorithm 04 : 尋找兩個有序陣列中的第N個數，要求時間複雜度為O(logm+logn)

Question : Give a divide and conquer algorithm for the following problem : you are
given two sorted lists of size m and n, and are allowed unit time access
to the ith element of each list. Given an O(logm+logn) time algorithm for
computing the kth largest element in the union of the two lists.
(For simplicity, you can assume that the elements of the two lists are
distinct).

問題：尋找兩個有序陣列中的第N個數，要求時間複雜度為O(logm+logn)。

/**
  * @author YuHuang
  * @vision 2011-10-04
  * This program is only for algorithm training.
  *
  */

import java.lang.RuntimeException;

public class SearchKthNumber {

        private int[] aArray;
        private int[] bArray;

        public SearchKthNumber(int[] aArray,int[] bArray){
                this.aArray=aArray;
                this.bArray=bArray;
                if(this.aArray==null||this.bArray==null){
                        throw new RuntimeException("Array should not be null!");
                }
        }

        public int doSearch(int aLeft,int aRight,int bLeft,int bRight,int k){
                int aMiddle = (aLeft+aRight)/2;
                int bMiddle = (bLeft+bRight)/2;

                if(aLeft>aRight){
                        return bArray[bLeft+k-1];
                }

                if(bLeft>bRight){
                        return aArray[aLeft+k-1];
                }

                if(aArray[aMiddle]>bArray[bMiddle]){
                        if(k<=(aMiddle-aLeft)+(bMiddle-bLeft)+1){
                                return doSearch(aLeft,aMiddle-1,bLeft,bRight,k);
                        }else{
                                return doSearch(aLeft,aRight,bMiddle+1,bRight,k-(bMiddle-bLeft)-1);
                        }
                }else{
                        if(k<=((aMiddle-aLeft)+(bMiddle-bLeft)+1)){
                                return doSearch(aLeft,aRight,bLeft,bMiddle-1,k);
                        }else{
                                return doSearch(aMiddle+1,aRight,bLeft,bRight,k-(aMiddle-aLeft)-1);
                        }
                }
        }


        public static void main(String[] args){
                int[] aArray=new int[]{1,3,5,6,8,9,11,18,20};
                int[] bArray=new int[]{2,4,7,16,22,29,39,40,55,100};

                SearchKthNumber sn=new SearchKthNumber(aArray,bArray);
                int k=2;
                int result=sn.doSearch(0,aArray.length-1,0,bArray.length-1,k);
                System.out.println("The "+k+"th"+" Number : "+result);
        }

}

[color=blue]歡迎提出更優化的解決方案，謝謝！[/color]