Median of Two Sorted Arrays

题目：There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

leetcode将这个问题转换为求第k个元素，k的值为(A's length + B' Length)/2。

如果其中一个为空，在第k个是不为空的第k个元素；如果k为0，则为a或b的

较小的第一个元素。

public static double findMedianSortedArrays(int A[], int B[]) {    int m = A.length;    int n = B.length;    if ((m + n) % 2 != 0) // odd    return (double) findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1);    else { // even    return (findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1)     + findKth(A, B, (m + n) / 2 - 1, 0, m - 1, 0, n - 1)) * 0.5;    }    }    public static int findKth(int A[], int B[], int k,     int aStart, int aEnd, int bStart, int bEnd) {    int aLen = aEnd - aStart + 1;    int bLen = bEnd - bStart + 1;    // Handle special cases    if (aLen == 0)    return B[bStart + k];    if (bLen == 0)    return A[aStart + k];    if (k == 0)    return A[aStart] < B[bStart] ? A[aStart] : B[bStart];    int aMid = aLen * k / (aLen + bLen); // a's middle count    int bMid = k - aMid - 1; // b's middle count    // make aMid and bMid to be array index    aMid = aMid + aStart;    bMid = bMid + bStart;    if (A[aMid] > B[bMid]) {    k = k - (bMid - bStart + 1);    aEnd = aMid;    bStart = bMid + 1;    } else {    k = k - (aMid - aStart + 1);    bEnd = bMid;    aStart = aMid + 1;    }    return findKth(A, B, k, aStart, aEnd, bStart, bEnd);    }