从两个排序的数组中打印不常见的元素
给定两个排序的数组，输出应显示其不常见的元素
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<n1 and j<n2
IF array1[i]<array2[j]
Print array1[i++]
End IF
ELSE If array1[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示例
#include <stdio.h>
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] <array2[j]) //checking whether the element of array1 is smaller than array2
printf("%d\n", array1[i++]);
else if (array1[i] > array2[j]) //checking whether the element of array2 is smaller than array1
printf("%d\n", 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\n", array1[i++]);
while(j < n2 && array2[j]!=array1[i]) //print remaining array1
printf("%d\n", array2[j++]);
return 0;
}输出结果
如果我们运行上面的程序，那么它将生成以下输出
1 2 6 7 8 10 12