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)).
题意：寻觅两个有序数组的中位数，要求复杂度为O(log
(m+n)).
思路：问题本质其实就是求两个有序数组的第Kth的数，那末我们可以这样斟酌，分别求出a,b两个数组中第k/2th的数，这两个数有3种情况
当a[k/2]<b[k/2]时，那末原kth数肯定不在a[k/2]之前的数内，然后抛弃a[k/2]之前的所有数，再在剩余的数里求k-(k/2)th数，其余两种情况同理，递归2分，所以复杂度降到对数级别
double find_kth(int a[],int m,int b[],int n,int k){
if(m>n)
return find_kth(b,n,a,m,k);
if(m==0)
return b[k⑴];
if(k==1)
return min(a[0],b[0]);
int pa=min(k/2,m),pb=k-pa;
if(a[pa⑴]<b[pb⑴])
return find_kth(a+pa,m-pa,b,n,k-pa);
else if(a[pa⑴]>b[pb⑴])
return find_kth(a,m,b+pb,n-pb,k-pb);
else
return a[pa⑴];
}
class Solution {
public:
double findMedianSortedArrays(int A[], int m, int B[], int n) {
int sum=m+n;
if(sum%2){
return find_kth(A,m,B,n,sum/2+1);
}
else
return (find_kth(A,m,B,n,sum/2)+find_kth(A,m,B,n,sum/2+1))/2;
}
};