[LeetCode] #4 Median of Two Sorted Arrays

There are two sorted arrays A and B 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)).

本文利用二分查找的思想，如果A[k/2]<B[k/2]，那么A中前k/2个元素一定小于AB混合数组的第k小的元素，所以下次只需要在A剩下的元素和B中的元素中找第（k-k/2）小的元素即可。利用迭代算法求解。其中奇数长度数组中位数只有一个，偶数长度数组中位数为其中的两个数字的平均。时间：54ms

代码如下：

class Solution {
public:
    double findKSorteArrays(int A[], int m, int B[], int n, int k){
        if (m>n)
            return findKSorteArrays(B, n, A, m, k);
        if (m == 0)
            return B[k-1];
        if (k == 1)
            return min(A[0], B[0]);
        int px = min(k / 2, m), py = k - px;
        if (A[px-1] < B[py-1])
            return findKSorteArrays(A + px, m - px, B, n, k - px);
        else if (A[px-1] > B[py-1])
            return findKSorteArrays(A, m, B + py, n - py, k - py);
        else
            return A[px-1];
    }
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        int total = n + m;
        if (total & 0x1)
            return findKSorteArrays(A, m, B, n, total / 2 + 1);
        else
            return (findKSorteArrays(A, m, B, n, total / 2) + findKSorteArrays(A, m, B, n, total / 2+1))/2;
    }
};