查找平均值，中位数不使用数组

Question

我如何找到平均值，一组数字的中位数而不使用C中的数组？ 问题不是找到平均数或中位数的方法，而是如何存储一组数字并对它们执行一些操作（如果不允许使用数组）？

Answer 1

一个有趣的问题。

寻找一个地方来存储所有数字的关键。代码可以使用文件，链接列表等。下面使用链表和递归。

当count是偶数时，将其留给OP以便对多种模式或中位数进行优化。

typedef struct num_S { 
    struct num_S *prev; 
    double x; 
} num_T; 

void GetNum(num_T *prev) { 
    num_T current; 
    current.prev = prev; 
    // If new number found ... 
    if (scanf("%lf", &current.x) == 1) { 
    num_T *p = &current; 
    // Sort new entry into list 
    while (p->prev != NULL && p->x < p->prev->x) { 
     double t = p->x; 
     p->x = p->prev->x; 
     p->prev->x = t; 
     p = p->prev; 
    } 
    GetNum(&current); 
    // End of list -now process the list 
    } else { 
    unsigned ModeCount = 0; 
    double Mode = 0.0; 
    unsigned ModeCandidateCount = 0; 
    double ModeCandidate = 0.0/0.0; 
    unsigned Count = 0; 
    double SumX = 0.0; 
    double SumXX = 0.0; 
    num_T *p = current.prev; 
    while (p != NULL) { 
     Count++; 
     SumX += p->x; 
     SumXX += p->x * p->x; 
     if (p->x == ModeCandidate) { 
     ModeCandidateCount++; 
     } else { 
     ModeCandidateCount = 1; 
     ModeCandidate = p->x; 
     } 
     if (ModeCandidateCount > ModeCount) { 
     ModeCount = ModeCandidateCount; 
     Mode = p->x; 
     } 
     p = p->prev; 
    } 

    printf("Count = %u\n", Count); 
    if (Count > 0) { 
     printf("Mode = %lf Count %u\n", Mode, ModeCount); 
     printf("Mean = %lf\n", SumX/Count); 
     printf("STD = %lf\n", sqrt(Count * SumX - SumXX)/Count); 

     Count /= 2; 
     num_T *p = current.prev; 
     while (Count-- > 0) { 
     p = p->prev; 
     } 
     printf("Median = %lf\n", p->x); 
    } 
    fflush(stdout); 
    } 
} 

int main(void) { 
    GetNum(NULL); 
    return 0; 
} 

输入4 3 4 2 4 1 EOF
输出：

Count = 6 
Mode = 4.000000 Count 3 
Mean = 3.000000 
STD = 1.130388 
Median = 3.000000 

STD编号：STD Rapid calculation methods