多数组第K小数

给定两个有序数组arr1和arr2，在给定一个整数k，返回两个数组的所有数中第K小的数。

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；

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}))。

直接上代码：

class Solution {
public:
int getupmedian(vector<int> arr1,int start1,int end1, vector<int> arr2,int start2,int end2){
int mid1=(start1+end1)/2;
int mid2=(start2+end2)/2;
int offest=((end1-start1+1)&1)^1;
if(end1<start1)
return -1;
if(start1==end1)
return arr1[start1]<=arr2[start2]?arr1[start1]:arr2[start2];
if(arr1[mid1]==arr2[mid2])
return arr1[mid1];
else if(arr1[mid1]>arr2[mid2])
return getupmedian(arr1,start1,mid1,arr2,mid2+offest,end2);
else return getupmedian(arr1,mid1+offest,end1,arr2,start2,mid2);
}
int findKthNum(vector<int> arr1, vector<int> arr2, int kth) {
int len1=arr1.size(),len2=arr2.size();
if(kth<1 || kth>len1+len2)
return -1;
if(len1<len2)
arr1.swap(arr2);
len1=arr1.size();
len2=arr2.size();
if(kth<=len2)
return getupmedian( arr2,0,kth-1,arr1,0,kth-1);
if(kth>len1){
if(arr2[kth-len1-1]>=arr1[len1-1])
return arr2[kth-len1-1];
if(arr1[kth-len2-1]>=arr2[len2-1])
return arr1[kth-len2-1];
return getupmedian(arr2,kth-len1,len2-1,arr1,kth-len2,len1-1);
}
if(arr1[kth-len2-1]>=arr2[len2-1])
return arr1[kth-len2-1];
return getupmedian( arr2,0,len2-1,arr1,kth-len2,kth-1);
}
};