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)).

 

class Solution {public:    int FindKthElm(int A[],int bA,int eA,int B[],int bB,int eB,int k){        if(bA>eA){            return B[bB+k-1];            }        if(bB>eB){            return A[bA+k-1];        }        int mA=(bA+eA)/2;        int mB=(bB+eB)/2;        int half=mA-bA+mB-bB+2;        if(A[mA]>B[mB]){ //此时B数组和第k元素都在A[mA]左侧，A[mA]到A[eA]可以舍弃            if(k
     
      half){				return FindKthElm(A,mA+1,eA,B,mB+1,eB,k-(half));			}			else return A[mA];		}        else{ //此时A数组和第k元素都在B[mB]左侧，B[mB]到B[eB]可以舍弃            if(k
     

 

public class Solution {   public int FindKth(int A[],int B[],int sa,int ea,int sb,int eb,int k){        if(sa>ea){            return B[sb+k-1];        }        else if(sb>eb){            return A[sa+k-1];        }        else{            int midA=(sa+ea)/2;            int midB=(sb+eb)/2;            int half=midA-sa+midB-sb+2;            if(A[midA]>B[midB]){                if(k
      
       half){                    k-=half;                    return FindKth(A,B,midA+1,ea,midB+1,eb,k);                }                else if(k