[Leetcode] Median of Two Sorted Arrays

大致思路：

先寻找每个数组的中位数，因为是排好顺序的数组，因此，可以在O(1)时间内找到。然后，比较这两个数字的大小。如果A的中位数大于B的中位数，则在A的前半个数组和B的后半个数组中寻找; 反之，在B的前半个数组和A的后半个数组寻找。根据递归方程，解得时间复杂度是O(lg(n)).

A general version:

Find the Kth element of two sorted arrays A[] and B[].

#include <cstdio>
#include <iostream>
#include <cstdlib>
using namespace std;

int findKthElem(int A[], int aBeg, int aEnd, int B[], int bBeg, int bEnd, int k){

    //cout << aBeg << " | " << aEnd << " | " << bBeg << " | " << bEnd << " | " << k << endl;

    if(aBeg > aEnd){
        return B[bBeg + k - 1];
    }
    if(bBeg > bEnd){
        return A[aBeg + k - 1];
    }

    int aMid = (aBeg + aEnd) >> 1;
    int bMid = (bBeg + bEnd) >> 1;
    int len = aMid - aBeg + bMid - bBeg + 2;

    if(len > k){
        if(A[aMid] < B[bMid]){
            return findKthElem(A, aBeg, aEnd, B, bBeg, bMid-1, k);
        }
        else{
            return findKthElem(A, aBeg, aMid-1, B, bBeg, bEnd, k);
        }
    }
    else{
        if(A[aMid] < B[bMid]){
            return findKthElem(A, aMid+1, aEnd, B, bBeg, bEnd, k-(aMid-aBeg+1));
        }
        else{
            return findKthElem(A, aBeg, aEnd, B, bMid+1, bEnd, k-(bMid-bBeg+1));
        }
    }
}

int main()
{
    const int ALen = 11;
    const int BLen = 5;

    int apos = 0;
    int bpos = 0;
    int A[ALen];
    int B[ALen];

    //生成两个递增数组A 和 B
    for (int i = 1; i <= ALen + BLen; ++i)
    {
        if (apos >= ALen)
        {
            B[bpos++] = i;
        }
        else if (bpos >= BLen)
        {
            A[apos++] = i;
        }
        else
        {
            if (rand()%2 == 1)
            {
                A[apos++] = i;
            }
            else
            {
                B[bpos++] = i;
            }
        }
    }

    //输出A和B的内容
    for (int i = 0; i < ALen; ++i)
    {
        cout <<A[i] <<" ";
    }
    cout <<endl;
    for (int i = 0; i < BLen; ++i)
    {
        cout <<B[i] <<" ";
    }
    cout <<endl;

    //验证每个K是不是正解
    for (int i = 1; i <= ALen + BLen; ++i)
    {
        cout << i <<" : "<<findKthElem(A, 0 , ALen - 1, B, 0 , BLen - 1, i)<<endl;
    }

    return 0;
}

参考：

http://blog.csdn.net/realxie/article/details/8078043

http://doudouclever.blog.163.com/blog/static/175112310201151584521268/

http://www.cnblogs.com/etcow/archive/2012/09/26/2704855.html