打印丢失的元素，范围在0 – 99之间
它将显示用户输入的给定集合中的缺失值
Given : array = {88, 105, 3, 2, 200, 0, 10};
Output : 1 4-9 11-87 89-99算法
START
STEP 1-> Take an array with elements, bool flag[MAX] to Fale, int i, j, n to size of array
Step 2-> Loop For from I to 0 and i<n and i++
IF array[i] < 100 && array[i]>=0
Set flag[array[i]]=true
End IF
Step 3 -> End For Loop
Step 4 -> Loop For from i to 0 and i<MAX and ++i
IF flag[i] == false
Set j=i+1
Loop While j<MAX && flag[j] == false
Set j++
End While
If j=i+1
Print i
End IF
Else
Print i and j-1
End Else
Set i=j
End IF
Step 5 -> End For Loop
STOP示例
#include <stdio.h>
#define MAX 100
int main(int argc, char const *argv[]) {
int array[] = {88, 105, 3, 2, 200, 0, 10};
bool flag[MAX] = { false }; //Initializing all the values of flag as false
int i, j, n;
n = sizeof(array)/sizeof(array[0]);
for (i = 0; i < n; i++) {
if (array[i] < 100 && array[i]>=0) {
flag[array[i]] = true; //Making the value of the elements present in an array as true, So missing will remain false
}
}
for (i = 0; i < MAX; ++i) {
if(flag[i] == false) { //Checking for false values
j = i+1; //Giving the value of the next iteration
while(j<MAX && flag[j] == false) //Checking the value of flag[j] is false
j++;
if (j==i+1) //For printing the missing number
printf("%d\n", i);
else //For printing the missing range
printf("%d-%d\n", i, j-1);
i = j; //Initializing the range's last value to start from that number
}
}
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 4-9 11-87 89-99