从两个已排序的数组中打印出不常见的元素

给定两个已排序的数组，输出应显示它们的非公共元素

Given : array1[]= {1, 4, 6, 9, 12}
   array2[]= {2, 4, 7, 8, 9, 10}
Output : 1 2 6 7 8 10 12

算法

START
Step 1 -> declare two arrays array1 and array2 with elements as int and variables n1, n2, i to 0 and j to 0
Step 2 -> calculate number of elements in array1 sizeof(array1)/sizeof(array1[0])
Step 3-> calculate number of elements in array2 sizeof(array2)/sizeof(array2[0])
Step 4 -> Loop While till i array2[j]
         Print array2[j++ ]
      End ELSE IF
   ELSE
      i++ and j++
   End ELSE
Step 5 -> End Loop While
Step 6 -> loop While i < n1 && array1[i]!=array2[j]
   Print array1[i++]
Step 7 -> End Loop While
Step 8 -> loop While j < n2 && array2[j]!=array1[i]
   Print array2[j++]
Step 9 -> End Loop While
STOP

Example

示例

#include 
int main(int argc, char const *argv[]) {
   int array1[]= {1, 4, 6, 9, 12};
   int array2[]= {2, 4, 7, 8, 9, 10};
   int n1, n2, i=0, j=0;
   n1 = sizeof(array1)/sizeof(array1[0]); //Calculating number of elements in array1
   n2 = sizeof(array2)/sizeof(array2[0]); //Calculating number of elements in array2
   while(i < n1 && j < n2) {
      if(array1[i] 
", array1[i++]);
      else if (array1[i] > array2[j]) //checking whether the element of array2 is smaller than array1
         printf("%d
", array2[j++]);
      else { //if they are equal increment both i and j      
         i++;
         j++;
      }
   }
   while(i < n1 && array1[i]!=array2[j]) //print remaining array1
      printf("%d
", array1[i++]);
   while(j < n2 && array2[j]!=array1[i]) //print remaining array1
      printf("%d
", array2[j++]);
   return 0;
}

输出

如果我们运行上面的程序，它将生成以下输出

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1
2
6
7
8
10
12