C++ 两个大小不同的排序数组的中值
我是从极客那里为极客解决这个问题的,但当我实现相同的代码时,我就开始为输入解决这个问题 A[]={1,2,3} B[]={3,6,9,12} 它给出了错误的输出为6,但它应该是3,有人请告诉我我的代码的问题C++ 两个大小不同的排序数组的中值,c++,arrays,binary-search,C++,Arrays,Binary Search,我是从极客那里为极客解决这个问题的,但当我实现相同的代码时,我就开始为输入解决这个问题 A[]={1,2,3} B[]={3,6,9,12} 它给出了错误的输出为6,但它应该是3,有人请告诉我我的代码的问题 //code #include<iostream> #include<cmath> using namespace std; float single_median(int arr[],int size)//method to find median in a
//code
#include<iostream>
#include<cmath>
using namespace std;
float single_median(int arr[],int size)//method to find median in an arr
{
if(size == 0)
return -1;
else if(size%2==0)
return (arr[size/2] + arr[size/2 -1])/2.0;
else
return arr[size/2];
}
float medianOf2(int a,int b)//median of two numbers
{
return ((a+b)/2.0);
}
int medianOf3(int a,int b,int c)//median of 3 numbers
{
int maximum = max(a,max(b,c));
int minimum = min(a,min(b,c));
return ((a+b+c) - maximum - minimum);
}
int medianOf4(int a, int b,int c,int d)//median of 4 numbers
{
int maximum = max(a,max(b,max(c,d)));
int minimum = min(a,min(b,min(c,d)));
return ((a+b+c+d) - maximum - minimum);
}
int find_median(int A[],int m,int B[],int n)
{
if(m<n)//here we will keep in mind that A is larger than B else we swap
return find_median(B,n,A,m);
if(n==0)//if smaller array has no element just find the median of larger array
return single_median(A,m);
if(n==1)//if smaller array has one element
{
if(m==1)
return (A[0]+B[0])/2.0;//if both has one element just return the average
else if(m&1)//when larger array has odd elements
return medianOf2(medianOf3(B[0],A[m/2 - 1],A[m/2 + 1]),A[m/2]);
else//for e
return medianOf3(B[0],A[m/2],A[m/2 -1]);
}
if(n==2)
{
if(m==2)
return medianOf4(A[0],B[0],A[1],B[1]);
else if(m&1)
return medianOf3(max(B[0],A[m/2 -1]),min(B[1],A[m/2 +1]),A[m/2]);
else
return medianOf4(max(B[0],A[m/2 -2]),min(B[1],A[m/2 +1]),A[m/2],A[m/2 -1]);
}
int mid_m = (m-1)/2;
int mid_n = (n-1)/2;
if(A[mid_m]<B[mid_n])
find_median(A + mid_m,m/2 +1 ,B,n - mid_n);
else
find_median(A,n/2 +1, B + mid_n, n/2 + 1);
}
int main()
{
int B[] = {1,2,3};
int A[] = {3,6,9,12};
cout<<find_median(A,4,B,3);
return 0;
}
//代码
#包括
#包括
使用名称空间std;
float single_median(int-arr[],int-size)//在arr中查找中值的方法
{
如果(大小==0)
返回-1;
else if(大小%2==0)
返回(arr[size/2]+arr[size/2-1])/2.0;
其他的
返回arr[size/2];
}
float medianOf2(inta,intb)//两个数字的中间值
{
回报率((a+b)/2.0);
}
int medianOf3(int a,int b,int c)//3个数字的中间值
{
int max=max(a,max(b,c));
最小整数=最小(a,最小(b,c));
返回((a+b+c)-最大值-最小值);
}
int medianOf4(int a,int b,int c,int d)//4个数字的中间值
{
int max=max(a,max(b,max(c,d));
最小整数=min(a,min(b,min(c,d));
返回((a+b+c+d)-最大-最小);
}
int find_中位数(int A[],int m,int B[],int n)
{
如果(m在返回结果之前忘记在medianOf4
函数中将结果除以2.0:
int medianOf4(int a, int b,int c,int d)//median of 4 numbers
{
int maximum = max(a,max(b,max(c,d)));
int minimum = min(a,min(b,min(c,d)));
return ((a+b+c+d) - maximum - minimum) / 2.0; # <-- forgot to add "/ 2.0"
}
intmedianof4(inta,intb,intc,intd)//4个数字的中间值
{
int max=max(a,max(b,max(c,d));
最小整数=min(a,min(b,min(c,d));
返回值((a+b+c+d)-最大值-最小值)/2.0;公共类解决方案{
公共双FindMediaTransortedArray(int[]nums1,int[]nums2){
如果此处不允许(nums1==null | | | nums1.length“告诉我代码的问题”之类的问题,请参阅。好的,先生,我不会重复此操作,谢谢通知。返回((a+b+c+d)-最大-最小);
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public class Solution {
public double FindMedianSortedArrays(int[] nums1, int[] nums2) {
if(nums1== null || nums1.Length<1)
return result(nums2);
else if(nums2== null || nums2.Length<1)
return result(nums1);
int[] merged = new int[nums1.Length + nums2.Length];
int num1Ind=0;
int num2Ind=0;
int currInd = 0;
while(num1Ind<nums1.Length && num2Ind<nums2.Length){
if(nums1[num1Ind]< nums2[num2Ind])
merged[currInd++]=nums1[num1Ind++];
else
merged[currInd++]=nums2[num2Ind++];
}
if(num2Ind<nums2.Length)
for(int i=num2Ind; i<nums2.Length; i++)
merged[currInd++]=nums2[i];
else if(num1Ind<nums1.Length)
for(int i=num1Ind; i<nums1.Length; i++)
merged[currInd++]=nums1[i];
return result(merged);
}
private double result(int[] merged){
if(merged.Length==1)
return merged[0];
if((merged.Length)%2 !=0)
return (double)merged[merged.Length/2];
return (double)(merged[merged.Length/2-1] + merged[merged.Length/2])/2;
}