【转载】【leetcode】Median of Two Sorted Arrays

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http://www.cnblogs.com/dollarzhaole/archive/2013/06/24/3153338.html

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)).

解题思路：

该题可以解决所有求有序数组A和B有序合并之后第k小的数！

该题的重要结论：

如果A[k/2-1]<B[k/2-1]，那么A[0]~A[k/2-1]一定在第k小的数的序列当中，可以用反证法证明。

具体的分析过程可以参考http://blog.csdn.net/zxzxy1988/article/details/8587244

class Solution {public:    double findKth(int A[], int m, int B[], int n, int k)    {        //m is equal or smaller than n        if (m > n)            return findKth(B, n, A, m, k);        if (m == 0)            return B[k-1];        if (k <= 1)            return min(A[0], B[0]);        int pa = min(k / 2, m), pb = k - pa;        if (A[pa-1] < B[pb-1])        {            return findKth(A + pa, m - pa, B, n, k - pa);        }        else if(A[pa-1] > B[pb-1])        {            return findKth(A, m, B + pb, n - pb, k - pb);        } else            return A[pa-1];    }    double findMedianSortedArrays(int A[], int m, int B[], int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int k = m + n;        if (k & 0x1)        {            return findKth(A, m, B, n, k / 2 + 1);        } else        {            return (findKth(A, m, B, n, k / 2) + findKth(A, m, B, n, k / 2 + 1)) / 2;        }    }};