**Leetcode_median-of-two-sorted-arrays (c++ and python version)

地址:http://oj.leetcode.com/problems/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)).

思路：这题参考了github上一份代码，注意时间复杂度要求是O(log (m+n))，看到这个应该联想到是系统递归的栈消耗。

find_kth_num(第K小数)使用于在两个array里找到第K小的数，不仅仅是中值。

另外注意k和下标索引差1. 

这道题刚开始做比较难理解，需要多揣摩。

c++ 参考代码;

class Solution {public:    //find kth smallest number, which applies to any kth number    double find_kth_num(int A[], int m, int B[], int n, int k)    {        if(m > n)            return find_kth_num(B, n, A, m, k);        if(m==0)            return B[k-1];        if(k == 1)            return min(A[0], B[0]);        int idx_a = min(m, k/2);        int idx_b = k - idx_a;        if(A[idx_a-1] < B[idx_b-1])            return find_kth_num(A+idx_a, m-idx_a, B, n, k-idx_a);        else if(A[idx_a-1] > B[idx_b-1])            return find_kth_num(A, m, B+idx_b, n-idx_b, k-idx_b);        else        //these two parts together count to lenght k, just find the kth smallest number            return A[idx_a-1];            }    double findMedianSortedArrays(int A[], int m, int B[], int n) {        if((m+n)&1)            return (double)find_kth_num(A, m, B, n, (m+n)/2+1);        else            return 0.5 * (find_kth_num(A, m, B, n, (m+n)/2) + find_kth_num(A, m, B, n, (m+n)/2+1));    }};

python 参考代码：

class Solution:    # @return a float    def find_kth_num(self, A, m, B, n, k):        if m > n:            return self.find_kth_num(B, n, A, m, k)        if not A:            return float(B[k-1])        if k == 1:            return float(min(A[0], B[0]))        idx_a = min(m, k//2)        idx_b = k - idx_a        if A[idx_a-1] < B[idx_b-1]:            return self.find_kth_num(A[idx_a:], m-idx_a, B, n, k-idx_a)        elif A[idx_a-1] > B[idx_b-1]:            return self.find_kth_num(A, m, B[idx_b:], n-idx_b, k-idx_b)        else:            return float(A[idx_a-1])                def findMedianSortedArrays(self, A, B):        if (len(A)+len(B)) & 0x1:            return self.find_kth_num(A, len(A), B, len(B), (len(A)+len(B)+1)//2)        else:            return (self.find_kth_num(A, len(A), B, len(B), (len(A)+len(B))//2) + self.find_kth_num(A, len(A), B, len(B), (len(A)+len(B))//2 + 1)) / 2.0

c++

Another trial, more simpler

class Solution {private:    double find(int A[], int m, int B[], int n, int k) {        if(m > n)            return find(B, n, A, m, k);        if(!m)            return B[k-1];        if(k==1)            return min(A[0], B[0]);                    int x = min(k/2, m);        if(A[x-1] < B[x-1])            return find(A+x, m-x, B, n, k-x);        else            return find(A, m, B+x, n-x, k-x);    }public:    double findMedianSortedArrays(int A[], int m, int B[], int n) {        if((m+n)%2)            return find(A, m, B, n, (m+n)/2+1);        else            return 0.5*(find(A, m, B, n, (m+n)/2) + find(A, m, B, n, (m+n)/2+1));    }};