原文:
给定一个整数数组，打印 0-99 范围内缺失的元素。如果有一个以上的缺失，整理它们，否则只打印数字。 注意，输入数组可能没有排序，可能包含范围[0-99]之外的数字，但打印缺失元素时只考虑这个范围。
示例:
input: {88, 105, 3, 2, 200, 0, 10}
output: 1
4-9
11-87
89-99
input: {9, 6, 900, 850, 5, 90, 100, 99}
output: 0-4
7-8
10-89
91-98
预期时间复杂度 o(n)，其中 n 是输入数组的大小。
想法是使用大小为 100 的布尔数组来跟踪 0 到 99 范围内的数组元素。我们首先遍历输入数组，并在布尔数组中标记这些当前元素。一旦标记了所有存在的元素，布尔数组就被用来打印丢失的元素。
以下是上述想法的实现。
c
// c program for print missing elements
#include
#define limit 100
using namespace std;
// a o(n) function to print missing elements in an array
void printmissing(int arr[], int n)
{
// initialize all number from 0 to 99 as not seen
bool seen[limit] = {false};
// mark present elements in range [0-99] as seen
for (int i=0; i
c
// c program for print missing elements
#include
#define limit 100
// a o(n) function to print missing elements in an array
void printmissing(int arr[], int n)
{
// initialize all number from 0 to 99 as not seen
bool seen[limit] = {false};
// mark present elements in range [0-99] as seen
for (int i=0; i
java 语言(一种计算机语言，尤用于创建网站)
class printmissingelement
{
// a o(n) function to print missing elements in an array
void printmissing(int arr[], int n)
{
int limit = 100;
boolean seen[] = new boolean[limit];
// initialize all number from 0 to 99 as not seen
for (int i = 0; i < limit; i )
seen[i] = false;
// mark present elements in range [0-99] as seen
for (int i = 0; i < n; i )
{
if (arr[i] < limit)
seen[arr[i]] = true;
}
// print missing element
int i = 0;
while (i < limit)
{
// if i is missing
if (seen[i] == false)
{
// find if there are more missing elements after i
int j = i 1;
while (j < limit && seen[j] == false)
j ;
// print missing single or range
int p = j-1;
system.out.println(i 1==j ? i : i "-" p);
// update u
i = j;
}
else
i ;
}
}
// driver program to test above functions
public static void main(string[] args)
{
printmissingelement missing = new printmissingelement();
int arr[] = {88, 105, 3, 2, 200, 0, 10};
int n = arr.length;
missing.printmissing(arr, n);
}
}
python 3
# python3 program for print missing elements
# a o(n) function to print missing elements in an array
def printmissing(arr, n) :
limit = 100
seen = [false]*limit
# initialize all number from 0 to 99 as not seen
for i in range(limit) :
seen[i] = false
# mark present elements in range [0-99] as seen
for i in range(n) :
if (arr[i] < limit) :
seen[arr[i]] = true
# print missing element
i = 0
while (i < limit) :
# if i is missing
if (seen[i] == false) :
# find if there are more missing elements after i
j = i 1
while (j < limit and seen[j] == false) :
j = 1
# print missing single or range
p = j - 1
if(i 1 == j) :
print(i)
else :
print(i, "-", p)
# update u
i = j
else :
i = 1
# driver code
arr = [88, 105, 3, 2, 200, 0, 10]
n = len(arr)
printmissing(arr, n)
# this code is contributed by divyesh072019.
c
using system;
class gfg
{
// a o(n) function to print missing elements in an array
static void printmissing(int[] arr, int n)
{
int limit = 100;
bool[] seen = new bool[limit];
int i;
// initialize all number from 0 to 99 as not seen
for (i = 0; i < limit; i )
seen[i] = false;
// mark present elements in range [0-99] as seen
for (i = 0; i < n; i )
{
if (arr[i] < limit)
seen[arr[i]] = true;
}
// print missing element
i = 0;
while (i < limit)
{
// if i is missing
if (seen[i] == false)
{
// find if there are more missing elements after i
int j = i 1;
while (j < limit && seen[j] == false)
j ;
// print missing single or range
int p = j - 1;
if(i 1 == j)
{
console.writeline(i);
}
else
{
console.writeline(i "-" p);
}
// update u
i = j;
}
else
i ;
}
}
// driver code
static void main()
{
int[] arr = {88, 105, 3, 2, 200, 0, 10};
int n = arr.length;
printmissing(arr, n);
}
}
// this code is contributed by divyeshrabadiya07.
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
输出:
1
4-9
11-87
89-99
上述程序的时间复杂度为 o(n)。
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