Question

int A[5] = {2, 2, 1, 4, 1};    
int max = A[0], min = A[0];
int k,l;

for (i = 0; i < 5; i++) {
    if (A[i] >= max) {
        max = A[i];
        k = i;
    }
}

for (j = 0; j < 5; j++) {
    if (A[j] <= min) {
        min = A[j];
        l = j;
    }
}

printf("densely is: %d ", k);
printf("\n");
printf("loosely is: %d ", l);

程序会将densely（max）的位置打印为3。它将loosely（min）的位置打印为4。但loosely（min）职位的正确答案应为2和4。就我而言，max号码为4，因此排名应为3（从0开始计算）。 min号码为1，因此排名应为2和4。现在，我的代码只显示min号码的一个位置：4。它应该打印两个位置：2和4。

我想打印max和min个数字的所有位置。

Answer 1

这实际上是非常好的问题。解决方案的关键是在数组min中找到max和A值。这部分已经完成并运作良好。这里唯一的改进是从索引1开始，从=和>=比较中删除<=。缺少的部分是记住min和max的所有索引。

一次通过很难做到。您可以一次性完成的工作是查找min和max。

在第二次传递中有min和max之后，您可以在名为max的其他索引表中标记min和maximums的所有索引位置和minimums。（因为我想打印这些值，我有两个独立的循环。）

解决方案非常简单：

#include <stdio.h>

// Find all minimums and all maximums in the array A
// Find and remember where `max` and `min` values were in the array A
// The algorithm is symmetrical for minimum and maximum 

int main(void)
{             // 0  1  2  3  4 
    int A[5] = { 2, 2, 1, 4, 1};    

    int maximums[5] = {0};  // we keep max positions here
    int minimums[5] = {0};  // we keep min positions here

    int max = A[0];  // we assume max at index 0
    int min = A[0];  // we assume min at index 0

    for (int i = 1; i < 5; i++) {       // we can start from index 1 because of the above assumption
        if (A[i] > max) max = A[i];     // densely 
        if (A[i] < min) min = A[i];     // loosely
    }

    // Find the all elements equal to `max` and `min` 
    // and register them in `maximums` and `minimums` arrays

    // Print the results:
    printf("max value is: %d and it occurs at the following index(es):\n", max);         
    for (int i = 0; i < 5; i++) {

        if (A[i] == max){ 
            maximums[i] = 1;   // mark this index as maximum being here
            printf("%d ", i);
        }
    }

    printf("\nmin value is: %d and it occurs at the following index(es):\n", min );   
    for (int i = 0; i < 5; i++) {

        if (A[i] == min){ 
            minimums[i] = 1;   // mark this index as minimum being here
            printf("%d ", i);
        }    
    }

    // At this moment we not only printed the results 
    // but the array minimums and maximums remember where the `min` and `max` were.

    return 0;
}

输出：

max value is: 4 and it occurs at the following index(es):                                                                                       
3                                                                                                                                               
min value is: 1 and it occurs at the following index(es):                                                                                       
2 4

如何在数组中找到最大和最小数字的位置？

1 个答案: