c语言++数组名【数字】

Problem statement: Write a C++ program to print all the non-repeated numbers in an array in minimum time complexity.

问题陈述：编写一个C ++程序， 以最小的时间复杂度将所有未重复的数字打印在数组中 。

Input Example:

输入示例：

Array length: 10    Array input: 2 5 3 2 4 5 3 6 7 3    Output:    Non-repeated numbers are: 7, 6, 4

Solution

解

Data structures used:

使用的数据结构：

Unordered_map 
    

• Key in the map is array value

  映射中的键是数组值

• Value of key is frequency

  关键值是频率

Algorithm:

算法：

1. Declare a map hash to store array elements as keys and to associate their frequencies with them.

   声明地图哈希，以将数组元素存储为键并将其频率与它们关联。

2. Unordered_map 
         
          hash;
         

3. For each array element

   对于每个数组元素

   Insert it as key & increase frequencies. (0 ->1)

   将其作为键插入并增加频率。 (0-> 1)

   For same key it will only increase frequencies.

   对于相同的键，只会增加频率。

4. For i=0: n-1	hash[array [i]]++;End For

5. Now to print the non-repeated character we need to print the keys (array elements) having value (frequency) exactly 1. (Non-repeating)

   现在要打印非重复字符，我们需要打印值(频率)正好为1的键(数组元素)。(非重复)

   Set an iterator to

   将迭代器设置为

   hash.begin().

   hash.begin() 。

   iterator->first is the key (array element) & iterator->second is the value( frequency of corresponding array value)

   iterator-> first是键(数组元素)＆ iterator-> second是值(对应数组值的频率)

6. IF    Iterator->second > 1    Print iterator->first (the array element)END IF

Time complexity: O(n)

时间复杂度：O(n)

Explanation with example:

举例说明：

For this array: 2 5 3 2 4 5 3 6 7 3

对于此阵列： 2 5 3 2 4 5 3 6 7 3

The code:

代码：

for(int i=0;i

Actually does the following

实际上是以下

At i=0    array[i]=2    Insert 2 & increase frequency    Hash:    Key(element)	Value(frequency)    2	            1    At i=1    array[i]=5    Insert 5 & increase frequency    Hash:    Key(element)	Value(frequency)    2	            1    5	            1    At i=2    array[i]=3    Insert 3 & increase frequency    Hash:    Key(element)	Value(frequency)    2	            1    5	            1    3	            1    At i=3    array[i]=2    Insert 2 increase frequency    '2' is already there, thus frequency increase.    Hash:    Key(element)	Value(frequency)    2	            2    5	            1    3	            1    At i=4    array[i]=4    Insert 4 &increase frequency    Hash:    Key(element)	Value(frequency)    2	            2    5	            1    3	            1    4	            1    At i=5    array[i]=5    '5' is already there, thus frequency increase.    Hash:    Key(element)	Value(frequency)    2	            2    5	            2    3	            1    4	            1    At i=6    array[i]=3    '3' is already there, thus frequency increase.    Hash:    Key(element)	Value(frequency)    2	            2    5	            2    3	            2    4	            1    At i=7    array[i]=6    Insert 6, increase frequency.    Hash:    Key(element)	Value(frequency)    2	            2    5	            2    3	            2    4	            1    6	            1    At i=8    array[i]=7    Insert 7, increase frequency.    Hash:    Key(element)	Value(frequency)    2	            2    5	            2    3	            2    4	            1    6	            1    7	            1    At i=9    array[i]=3    '3' is already there, thus frequency increase.    Hash:    Key(element)	Value(frequency)    2	            2    5	            2    3	            3    4	            1    6	            1    7	            1

Thus, Elements with frequency 1 are: 7, 6, 4

因此，频率为1的元素为：7、6、4

C ++实现可在数组中按频率打印所有非重复数字 (C++ implementation to print all the Non-Repeated Numbers with Frequency in an Array)

#include 
    
     using namespace std;void findNonRepeat(int* a, int n){
       //Declare the map    unordered_map
     
       hash;        for(int i=0;i
      
       first == key(element value)    //iterator->second == value(frequency)        for(auto it=hash.begin();it!=hash.end();it++)    if(it->second==1)//frequency==1 means non-repeating element    printf("%d ",it->first);        printf("\n");    }int main(){
       int n;    cout<<"enter array length\n";    cin>>n;    int* a=(int*)(malloc(sizeof(int)*n));        cout<<"input array elements...\n";        for(int i=0;i
      
     
    

Output

输出量

enter array length10input array elements...2 5 3 2 4 5 3 6 7 3the nonrepeating numbers are: 7 6 4

翻译自:

c语言++数组名【数字】