获取三个数的中间值 宏

日常遇到较多的时间使用都是比较两个数的最大值 或者最小值，这两个处理过程很简单，都可以使用 宏MAX/MIN 来计算。

今天碰到获取三个数的中间值使用情况，花费些时间思考，将个人思考的两种方法进行总结记录。

• 方法一

  将三个数相加，然后减去三个数中的最大值和最小值，余下的就是中间值。

  #define MIN(a,b) (((a)<(b))?(a):(b))
  #define MAX(a,b) (((a)>(b))?(a):(b))
  ​
       
  #define MID_1(a,b,c) ((a) + (b) + (c) - MAX(MAX(a,b),c) - MIN(MIN(a,b),c))*

• 方法二

  采用简单的逻辑判断，依次进行abc 三者之间的关系判断，直至获取到中间值。如下为逻辑代码：

  if a > b 
  
  ​	if a c
  
  ​			
  
  ​			mid = b
  
  ​		else
  
  ​			
  
  ​			mid = c
  
  else
  
  ​	if a > c
  
  ​		
  
  ​		mid = a
  
  ​	else
  
  ​		
  
  ​		if b > c
  
  ​			
  
  ​			mid = c
  
  ​		else
  
  ​			
  
  ​			mid = b

abc 三个数中，每个数与其他数的关系都是大于/等于/小于 ，因此测试用例总共有3^3 = 27 种。

#include
​

#define MIN(a,b) (((a)<(b))?(a):(b))
#define MAX(a,b) (((a)>(b))?(a):(b))
​
     
#define MID_1(a,b,c) ((a) + (b) + (c) - MAX(MAX(a,b),c) - MIN(MIN(a,b),c))

#define MID_2(a,b,c) ((a)>(b)?((a)<(c)?(a):(b)>(c)?(b):(c)):((a)>(c)?(a):(b)>(c)?(c):(b)))
​
​
int main()
{
    int num[3] = {1,2,3};
    int count = 3;
    int i = 0, j = 0, k = 0;
    int mid1 = -1,mid2 = -1;
​
    for(i = 0;i  MID_1 = %d, MID_2 = %d, MID_1 %s MID_2n",num[i],num[j],num[k],mid1,mid2,mid1==mid2?"==":"!=");
            }
​
    return 0;
}

结果如下：

MId(1,1,3) => MID_1 = 1, MID_2 = 1, MID_1 == MID_2
MId(1,2,1) => MID_1 = 1, MID_2 = 1, MID_1 == MID_2
MId(1,2,2) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(1,2,3) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(1,3,1) => MID_1 = 1, MID_2 = 1, MID_1 == MID_2
MId(1,3,2) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(1,3,3) => MID_1 = 3, MID_2 = 3, MID_1 == MID_2
MId(2,1,1) => MID_1 = 1, MID_2 = 1, MID_1 == MID_2
MId(2,1,2) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(2,1,3) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(2,2,1) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(2,2,2) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(2,2,3) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(2,3,1) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(2,3,2) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(2,3,3) => MID_1 = 3, MID_2 = 3, MID_1 == MID_2
MId(3,1,1) => MID_1 = 1, MID_2 = 1, MID_1 == MID_2
MId(3,1,2) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(3,1,3) => MID_1 = 3, MID_2 = 3, MID_1 == MID_2
MId(3,2,1) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(3,2,2) => MID_1 = 2, MID_2 = 2, MID_1 == MID_2
MId(3,2,3) => MID_1 = 3, MID_2 = 3, MID_1 == MID_2
MId(3,3,1) => MID_1 = 3, MID_2 = 3, MID_1 == MID_2
MId(3,3,2) => MID_1 = 3, MID_2 = 3, MID_1 == MID_2
MId(3,3,3) => MID_1 = 3, MID_2 = 3, MID_1 == MID_2

​可以看到两种结果计算的中间值是一样，结果均正确。