合并两个数组并省略所有重复元素

时间：2013-11-21 15:23:43

标签： c arrays merge

我有2个数组。输入1和输入2.假设输入1具有{1,1,2,3,3,4,5}，输入2具有{4,2,6,7,8}。合并数组{1,1,2,3,3,4,5,4,2,6,7,8}

排序后，我的合并数组看起来像{1,1,2,2,3,3,4,4,5,6,7,8}

我的输出应该像{5,6,7,8}，因为它们是非重复元素。它应该既没有在第一次发生两次，也没有在第一次发生一次，在第二次发生一

我已将两个数组放入一个单独的数组中（两者合并）。对它进行排序并删除重复项。我得到的输出像{1,2,3,4,5,6,7,8}，但我不应该得到1,2,3,4，因为它们在合并数组中都发生了两次。

请帮我尽快完成该计划。提前谢谢。

我不能使用结构。那是我的任务。

#include<stdio.h>
#include<conio.h>
main()
{
    int a[10],b[10],c[10],i,j,n,n1,temp;
    clrscr();
    printf("enter the no of ele in array1\n");
    scanf("%d",&n);
    printf("enter 1st array elements\n");
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    printf("enter size of 2nd array");
    scanf("%d",&n1);
    printf("enter 2nd array elements\n");
    for(i=0;i<n;i++)
    {
        scanf("%d",&b[i]);
    }

    for(i=0;i<n;i++)
    {
        c[i]=a[i];
    }
    j=0;
    for(i=n;i<n+n1;i++)
    {
        c[i]=b[j];
          //    printf("\n2nd array values are %d",c[i]);
        j++;
    }
    for(i=0;i<n+n1;i++)
    {
        printf("%d\n",c[i]);
    }
    for(i=0;i<n-1;i++)
    {
        for(j=0;j<n-i-1;j++)
        {
            if(c[j]>c[j+1])
            {
                temp=c[j];
                c[j]=c[j+1];
                c[j+1]=temp;
            }
        }
    }
    for(j=0;j<n+n1;j++)
    {
        printf("sorted is %d\n",c[j]);
    }

    for(i=0;i<n+n1;i++)
    {
        if(c[i]==c[i+1])
        {
            temp=c[i];

         // printf("%d r d rep.ele\n",c[i]);
        }
        else
        printf("%d r d ele\n",c[i]);
    }

    getch();
}

6 个答案:

答案 0 :(得分：1)

让我们以不恰当但又有趣的方式解决这个问题：

#include <stdio.h>

#define MAXINPUT 10

void pack(const int* k, int n, int* table)
{
  for(int i = 0; i < n; ++i)
  {
    table[k[i]]++;
  }
}

void dump(const int* table)
{
  int flag = 0;
  for(int i = 0; i < MAXINPUT; ++i)
  {
    if(table[i] == 1)
    {
      printf("%s%d", (flag) ? ", " : "", i);
      flag = 1;
    }
  }
}

int c[MAXINPUT];

int main(void)
{
  const int a[] = { 1, 1, 2, 3, 3, 4, 5};
  const int b[] = { 4, 2, 6, 7, 8};
  pack(&a[0], sizeof(a)/sizeof(int), &c[0]);
  pack(&b[0], sizeof(b)/sizeof(int), &c[0]);
  dump(&c[0]);
  return 0;
}

答案 1 :(得分：0)

size_t i, non_repeating_index = 0;

// allocate memory to hold the new array
int * non_repeating_elements = calloc(sizeof(int), merged_array_length);

// check if the first element is unique
if (merged_array_length >= 2 && merged_array[0] != merged_array[1])
    non_repeating_elements[non_repeating_index++] = merged_array[0];

// for all other elemnts check if it equals the ones around it
for (i = 1; i < merged_array_length -1; i++) {
    if (merged_array[i-1] != merged_array[i] && merged_array[i] != merged_array[i + 1])
        non_repeating_elements[non_repeating_index++] = merged_array[i];
}

// check the last element
if (merged_array_length >= 3 && merged_array[i-1] != merged_array[i])
    non_repeating_elements[non_repeating_index++] = merged_array[i];

// resize the array to conserve memory
non_repeating_elements = realloc(non_repeating_elements, sizeof(*non_repeating_elements) * non_repeating_index);

答案 2 :(得分：0)

你快到了！不要试图花哨，只需创建一个包含两个int字段的结构：value和count。创建这些结构的数组并遍历排序列表。最后，遍历结构数组并选择相应计数为1的所有值。

有一些细节可以解决，比如你的结构数组应该有多大，但是嘿，你显然不怕编码。如果阵列比你需要的大，那么......只要它足够大就关心它？

答案 3 :(得分：0)

如果你已经对数组进行了排序和合并，那么它非常简单：只需与前一代产品进行比较即可。接班人。但是需要一些额外的逻辑才能超出范围。

#include <stdio.h>

int main(int argc, char** argv){
  const int sortedMergedArray[] = {1,1,2,2,3,3,4,4,5,6,7,8};
  const size_t mergedArrayLength = 12;
  for(size_t i=0; i<mergedArrayLength;i++){
    int predecessor = -1;
    int successor = -1;
    if(i >= 1) predecessor = sortedMergedArray[i-1];
    if(i <= mergedArrayLength-1) successor = sortedMergedArray[i+1];
    if(sortedMergedArray[i] != successor &&
        sortedMergedArray[i] != predecessor){
      printf("%d, ", sortedMergedArray[i]);
    }
  }
  return 0;
}

输出：

5, 6, 7, 8,

答案 4 :(得分：0)

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

#define NUMCMP(x,y) (((x) < (y)) ? -1 : ((x) > (y)) ? 1 : 0)

#define numcmp(type) int cmp_##type(const void *a, const void *b){\
    type x = *(type*)a;\
    type y = *(type*)b;\
    return NUMCMP(x,y);\
}

numcmp(int);//define function cmp_int

void print_array(int a[], size_t size){
    size_t i;
    printf("{");
    for(i=0;i<size;++i){
        printf("%d", a[i]);
        if(i != size-1)
            printf(",");
    }
    printf("}\n");
}

int main(void){
    int input1[] = {1,1,2,3,3,4,5};
    int input2[] = {4,2,6,7,8};
    size_t input1_size = sizeof(input1)/sizeof(*input1);
    size_t input2_size = sizeof(input2)/sizeof(*input2);

    printf("input1: ");
    print_array(input1, input1_size);
    printf("input2: ");
    print_array(input2, input2_size);

    size_t marge_array_size = input1_size + input2_size;
    int marge_array[marge_array_size];
    memcpy(marge_array, input1, input1_size*sizeof(*input1));
    memcpy(&marge_array[input1_size], input2, input2_size*sizeof(*input2));
    printf("marge array: ");
    print_array(marge_array, marge_array_size);
    qsort(marge_array, marge_array_size, sizeof(*marge_array), cmp_int);
    printf("marge array(after sort): ");
    print_array(marge_array, marge_array_size);

    size_t singles_size = marge_array_size;
    int *singles = malloc(singles_size * sizeof(*singles));
    size_t i;
    int sp;
    bool drop = false;
    for(sp=-1, i=0;i<marge_array_size;++i){
        if(!drop){
            if(sp==-1){
                singles[++sp] = marge_array[i];
                continue;
            }
            if(singles[sp] == marge_array[i]){
                --sp;
                drop = true;
                continue;
            }
        } else if(singles[sp+1] == marge_array[i])
            continue;
        drop = false;
        singles[++sp] = marge_array[i];
    }
    if(sp+1 != singles_size){
        singles_size = sp+1;
        singles = realloc(singles, singles_size*sizeof(*singles));
    }
    printf("marge array(single num): ");
    print_array(singles, singles_size);

    size_t uniques_size = marge_array_size;
    int *uniques = malloc(uniques_size * sizeof(*uniques));
    for(sp=-1, i=0;i<marge_array_size;++i){
        if(sp==-1){
            uniques[++sp] = marge_array[i];
            continue;
        }
        if(uniques[sp] == marge_array[i]){
            continue;
        }
        uniques[++sp] = marge_array[i];
    }
    if(sp+1 != uniques_size){
        uniques_size = sp+1;
        uniques = realloc(uniques, uniques_size*sizeof(*uniques));
    }
    printf("marge array(uniques num): ");
    print_array(uniques, uniques_size);
    //be free uniques and singles 
    return 0;
}

答案 5 :(得分：0)

    #include<stdio.h>
    int main()
  {
     int a[30],b[30];
     int n,k,i,j,p,c=0;
    printf("How many element in first array\n");
    scanf("%d",&n);
    printf("Enter elements");
     for(i=0;i<n;i++)
    {
       scanf("%d",&a[i]);                //FIRST ARRAY INPUT
    }


   printf("How many element in second array\n");
  scanf("%d",&k);
  printf("Enter elements");
  for(i=0;i<k;i++)
  {
    scanf("%d",&b[i]);                // SECOND ARRAY INPUT
  }
 printf("The new array is \n");
 for(i=n,j=0;i<(n+k);i++)
   {  
     a[i]=b[j++];                  // MERGE ARRAY B[30],IN A[30] BY EXTEND IT.
   }

 for(i=0;i<(n+k);i++)
   {
     printf("%d ",a[i]);        // PRINT THE MERGED ARRAY
   }
for(i=0,p=0;i<(n+k);i++)        // ELIMINATE DUPLICATE ELEMENTS 
 {
    if(a[i]!=NULL)              
    {

        for(j=i+1;j<(n+k);j++)
        {
            if(a[i]==a[j])       // IF ANY ELEMENTS  MATCH TO NEXT OTHERS ARRAY ELEMENTS THEN, THOSE ELEMENTS ARE ASSIGNED BY NULL.   
            {
                a[j]=NULL;       
            }
        }
        a[p++]=a[i];
        c++;               // COUNTING FOR SIZE OF NEW ARRAY
    }


 }
printf("\nAfter eliminate the duplicate elements,the new array is : \n");
         for(i=0;i<c;i++)
         {
          printf("%d ",a[i]);           // PRINT NEW ARRAY
         }
  return 0;

  }
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