原文:
给定一个大小为 n、的 arr[] ，任务是打印中所有唯一的元素。
如果数组中某个元素的频率为 1 ，则称该元素为唯一。
示例:
输入: arr[ ] = {1，1，2，2，3，4，5，5} 输出: 3 4 解释:由于 1，2，5 在数组中出现不止一次，所以不同的元素是 3 和 4。
输入: arr[ ] = {1，2，3，3，3，4，5，6} 输出: 1 2 4 5 6
方法:解决问题最简单的方法就是遍历数组arr【】只打印那些频率为 1 的元素。按照以下步骤解决问题:
- arr[] 并初始化一个变量，比如 cnt = 0，来计算当前数组元素的频率。
- 由于已经是，检查当前元素是否与前一个元素相同。如果发现是真的，那么更新 cnt = 1 。
- 否则，如果 cnt = 1，则打印元素。否则，继续。
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// function to print all unique
// elements present in a sorted array
void removeduplicates(int arr[], int n)
{
int i = 0;
// traverse the array
while (i < n) {
int cur = arr[i];
// stores frequency of
// the current element
int cnt = 0;
// iterate until end of the
// array is reached or current
// element is not the same as the
// previous element
while (i < n and cur == arr[i]) {
cnt ;
i ;
}
// if current element is unique
if (cnt == 1) {
cout << cur << " ";
}
}
}
// driver code
int main()
{
// given input
int arr[] = { 1, 3, 3, 5, 5, 6, 10 };
int n = 7;
// function call
removeduplicates(arr, n);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.io.*;
class gfg
{
// function to print all unique
// elements present in a sorted array
static void removeduplicates(int arr[], int n)
{
int i = 0;
// traverse the array
while (i < n) {
int cur = arr[i];
// stores frequency of
// the current element
int cnt = 0;
// iterate until end of the
// array is reached or current
// element is not the same as the
// previous element
while (i < n && cur == arr[i]) {
cnt ;
i ;
}
// if current element is unique
if (cnt == 1) {
system.out.print(cur " ");
}
}
}
// driver code
public static void main (string[] args)
{
// given input
int arr[] = { 1, 3, 3, 5, 5, 6, 10 };
int n = 7;
// function call
removeduplicates(arr, n);
}
}
// this code is contributed by potta lokesh
python 3
# function to print all unique
# elements present in a sorted array
def removeduplicates(arr, n):
i = 0
while i < n:
cur = arr[i]
# stores frequency of
# the current element
cnt = 0
# iterate until end of the
# array is reached or current
# element is not the same as the
# previous element
while i < n and cur == arr[i]:
cnt = 1
i = 1
if cnt == 1:
print(cur, end=" ")
# driver code
if __name__ == "__main__":
# given input
arr = [1, 3, 3, 5, 5, 6, 10]
n = 7
# function call
removeduplicates(arr, n)
# this code is contributed by kushagra bansal
c
// c# program for the above approach
using system;
class gfg {
// function to print all unique
// elements present in a sorted array
static void removeduplicates(int[] arr, int n)
{
int i = 0;
// traverse the array
while (i < n) {
int cur = arr[i];
// stores frequency of
// the current element
int cnt = 0;
// iterate until end of the
// array is reached or current
// element is not the same as the
// previous element
while (i < n && cur == arr[i]) {
cnt ;
i ;
}
// if current element is unique
if (cnt == 1) {
console.write(cur " ");
}
}
}
// driver code
public static void main()
{
// given input
int[] arr = { 1, 3, 3, 5, 5, 6, 10 };
int n = 7;
// function call
removeduplicates(arr, n);
}
}
// this code is contributed by rishavmahato348.
java 描述语言
output
1 6 10
时间复杂度:o(n) t5辅助空间:** o(1)
麻将胡了pg电子网站的版权属于：月萌api www.moonapi.com，转载请注明出处
