给定两个有序数组arr1和arr2，再给定一个整数k,返回两个数组中所有数中第k小的数。要求：如果arr1的长度为N，arr2的长度为M，时间复杂度请达到O(log(min{M, N}) )

例如：
arr1 = {1,2,3,4,5}
arr2 = {3,4,5}
k = 1;
因为1为所有数中最小的，所以返回1
arr1 = {1,2,3}
arr2 = {3,4,5,6}
k = 4
因为3为所有数中第4小的数，所以返回3

#include <iostream>
#include <assert.h> 
#include <vector>
using namespace std;int FindKthNum(int* arr1, int len1, int* arr2, int len2, int k)
{assert(k >= 0 && len1 + len2 >= k);if(len1 > len2){return FindKthNum(arr2, len2, arr1, len1, k);}if(len1 == 0)return arr2[k-1];if(k == 1)return min(arr1[0], arr2[0]);                          int range1 = min(k/2, len1), range2 = k - range1;// arr1[0,range1) + arr2[0,range2) 是两个数组的前k个数// 1.arr1[range1-1] == arr2[range2-1]，则这个两个数其中一个就是第k个// 2.arr1[range1-1] < arr2[range2-1]，则排除arr1[0,range1)这一部分// 3.arr1[range1-1] > arr2[range2-1]，则排除arr2[0,range2)这一部分if(arr1[range1-1] == arr2[range2-1])return min(arr1[range1-1], arr2[range2-1]);else if(arr1[range1-1] < arr2[range2-1])//去掉range1个return FindKthNum(arr1+range1, len1-range1, arr2, len2, k-range1);elsereturn FindKthNum(arr1, len1, arr2+range2, len2-range2, k-range2);}void test()
{int a1[5] = {1,2,3,4,5};int a2[3] = {3,4,5};cout<<FindKthNum(a1, 5, a2, 3, 1)<<endl;int a11[3] = {1,2,3};int a22[4] = {3,4,5,6};vector<int> arr11(a11, a11+3);vector<int> arr22(a22, a22+4);cout<<FindKthNum(a11, 3, a22, 4, 4)<<endl;
}