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c - 清除c中的数组数据

问题描述

我写了一个关于归并排序的程序。我对数组有一些问题。

第一行代表输入序列的数量。（现在我输入 2，表示有 2 个输入）

Input: 2
9,5,6,7,1,8,3
22,86,-5,8,66,9

输出：

1 3 5 6 7 8 9
-5 1 3 5 6 7 8 8 9 9 22 66 86

我如何在下一秒输出数组没有第一个输入值9,5,6,7,1,8,3只有第二个输入22,86,-5,8,66,9

意味着输出应该像

1 3 5 6 7 8 9
-5 8 9 22 66 86

这是我的代码：

#include <string.h>
#include <stdio.h>
#include <stdlib.h>

// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
    int i, j, k;
    int n1 = m - l + 1;
    int n2 =  r - m;

    /* create temp arrays */
    int L[n1], R[n2];

    /* Copy data to temp arrays L[] and R[] */
    for (i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for (j = 0; j < n2; j++)
        R[j] = arr[m + 1+ j];

    /* Merge the temp arrays back into arr[l..r]*/
    i = 0; // Initial index of first subarray
    j = 0; // Initial index of second subarray
    k = l; // Initial index of merged subarray
    while (i < n1 && j < n2)
    {
        if (L[i] <= R[j])
        {
            arr[k] = L[i];
            i++;
        }
        else
        {
            arr[k] = R[j];
            j++;
        }
        k++;
    }

    /* Copy the remaining elements of L[], if there
       are any */
    while (i < n1)
    {
        arr[k] = L[i];
        i++;
        k++;
    }

    /* Copy the remaining elements of R[], if there
       are any */
    while (j < n2)
    {
        arr[k] = R[j];
        j++;
        k++;
    }
}

/* l is for left index and r is right index of the
   sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
    if (l < r)
    {
        // Same as (l+r)/2, but avoids overflow for
        // large l and h
        int m = l+(r-l)/2;

        // Sort first and second halves
        mergeSort(arr, l, m);
        mergeSort(arr, m+1, r);

        merge(arr, l, m, r);
    }
}

/* UTILITY FUNCTIONS */
/* Function to print an array */
void printArray(int A[], int size)
{
    int i;
    for (i=0; i < size; i++)
        printf("%d ", A[i]);
    printf("\n");
}

int main()
{

    int limit,i;
    //char *tree_input = malloc(1000*sizeof(char));
    //int *array1 = malloc(1000 * sizeof(int));
    char tree_input[1000];
    int array1[1000]; //char also can?
    int j=0;
    char *pch;

    printf("input the limit\n");
    scanf("%d",&limit);

    for(i=0; i<limit; i++)
    {

        scanf("%s",tree_input);
        pch = strtok(tree_input, ",");

        while(pch!=NULL)
        {
            //printf("%s\n",pch);
            array1[j]=atoi(pch); //atoi it
            j++;
            pch = strtok (NULL, ",");
        }
        mergeSort(array1, 0, j - 1);
        printArray(array1, j);
    }

    return 0;
}

标签： carrays

解决方案

您需要j在处理每个输入后重置。

否则，您array1将拥有所有输入。

for(i=0; i<limit; i++)
{

    scanf("%s",tree_input);
    pch = strtok(tree_input, ",");

    while(pch!=NULL)
    {
        //printf("%s\n",pch);
        array1[j]=atoi(pch); //atoi it
        j++;
        pch = strtok (NULL, ",");
    }
    mergeSort(array1, 0, j - 1);
    printArray(array1, j);
    j=0; //Reset the j
}

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