INDIA
Output
INDA
****
*
std::map中的键，其值为出现次数。然后我遍历该映射并打印星号，并为每个字符减少映射中的值。
最初，我被要求不使用任何库。我提供了一个需要大量迭代的解决方案。对于每个字符，迭代完整的数组直到索引以查找先前的出现次数等等。 是否有更好的方法，例如更好的复杂度，比如更快的操作，可以实现这一目标？
map，因为使用某种数据结构来复制map的基本功能几乎是解决这个问题最合理的方法。int组成的数组，并在其中存储每个字符的计数。#include <iostream>
int main()
{
char input[100];
unsigned int input_length = 0;
char letters[100];
unsigned int num_of_letters = 0;
std::cin >> input;
while (input[input_length] != '\0')
{
input_length += 1;
}
//This array acts like a hash map.
unsigned int occurrences[256] = {0};
unsigned int max_occurrences = 1;
for (int i = 0; i < input_length; ++i)
{
if ((occurrences[static_cast<unsigned char>(input[i])] += 1) == 1)
{
std::cout<< " " << (letters[num_of_letters] = input[i]) << " ";
num_of_letters += 1;
}
if (occurrences[static_cast<unsigned char>(input[i])] > max_occurrences)
{
max_occurrences = occurrences[static_cast<unsigned char>(input[i])];
}
}
std::cout << std::endl;
for (int row = 1; row <= max_occurrences; ++row)
{
for (int i = 0; i < num_of_letters; ++i)
{
if (occurrences[static_cast<unsigned char>(letters[i])] >= row)
{
std::cout << " * ";
}
else
{
std::cout << " ";
}
}
std::cout << std::endl;
}
return 0;
}
void printAstr(std::string str){
int array[256] ;// assumining it is an ascii string
memset(array, 0, sizeof(array));
int astrCount = 0;
for(int i = 0; i < str.length()-1; i++){
array[(int) str[i]]++;
if(array[(int) str[i]] > 1) astrCount++;
}
std::cout << str << std::endl;
for(int i = 0; i < str.length()-1;i++) std::cout << "* ";
std::cout << std::endl;
while(astrCount != 0){
for(int i= 0; i< str.length() - 1;i++){
if(array[(int) str[i]] > 1){
std::cout << "* ";
array[(int) str[i]]--;
astrCount--;
}else{
std::cout << " ";
}
}
std::cout << std::endl;
}
}
非常简单，只需将所有值添加到数组中，然后打印它们出现的次数。
编辑：抱歉刚刚做了一些逻辑更改。现在这个代码可以正常运行。
é。 - M.M这里还有一个例子： 你可以看到它在这里工作。
#include <stdio.h>
int main()
{
int i,j=0,f=1;
char input[50]={'I','N','D','I','A','N','A','N'};
char letters[256]={0};
int counter[256]={0};
for(i=0;i<50;i++)
{
if(input[i])
counter[input[i]]++;
if(counter[input[i]]==1)
{
putchar(input[i]);
letters[j]=input[i];
j++;
}
}
putchar('\n');
while(f)
{
f=0;
for(i=0;i<j;i++)
if(counter[letters[i]])
{
putchar('*');
counter[letters[i]]--;
f=1;
}
else
{
putchar(' ');
}
putchar('\n');
}
return 0;
}
nullptr，这里 是现场演示和下面的代码：struct letter_count
{
char letter = '\0';
int count = 0;
};
int add(letter_count *begin, letter_count *end, char letter)
{
while (begin != end)
{
if (begin->letter == letter)
{
return ++begin->count;
}
else if (begin->letter == '\0')
{
std::cout << letter; // Print the first appearance of each char
++begin->letter = letter;
return ++begin->count;
}
++begin;
}
return 0;
}
int max (int a, int b)
{
return a > b ? a : b;
}
letter_count *buffer = nullptr;
auto testString = "supergalifragilisticoespialidoso";
int len = 0, index = 0, greater = 0;
while (testString[index++])
++len;
buffer = new letter_count[len];
for (index = 0; index < len; ++index)
greater = max(add(buffer, buffer + len, testString[index]), greater);
std::cout << '\n';
for (int count = 0; count < greater; ++count)
{
for (index = 0; buffer[index].letter && index < len; ++index)
std::cout << (count < buffer[index].count ? '*' : ' ');
std::cout << '\n';
}
delete [] buffer;
由于“不允许使用库”（除了<iostream>？），我避免使用std::pair<char, int>（它本来可以替代letter_count结构体），我们需要编写许多工具程序（如max和strlen）；上述程序的输出结果为：
supergaliftcod
**************
* ******* *
* *** *
* *
*
*
#include <iostream>
#include <string>
using namespace std;
int
main ()
{
string input = "";
string tempstring = "";
string process = "";
string output = "";
bool test = false;
cout << "Enter your word below: " << endl;
cin >> input;
for (unsigned int i = 0; i < input.length (); i++)
{ //for the traversed letter, traverse through subsequent letters
for (unsigned int z = i + 1; z < input.length (); z++)
{
//avoid analyzing nonsense characters
if (input[i] != '!')
{
if (input[i] == input[z])
{ //matched letter; replace with nonsense character
input[z] = '!';
test = true; //for string management later
}
}
}
if (test)
{
tempstring += input[i];
input[i] = '*';
test = false; //reset bool for subsequent loops
}
}
//remove garbage symbols and save to a processing string
for (unsigned int i = 0; i < input.size (); i++)
if (input[i] != '!')
process += input[i];
//create the modified output string
unsigned int temp = 0;
for (unsigned int i = 0; i < process.size (); i++)
if (process[i] == '*')
{ //replace asterisks with letters stored in tempstring
output += tempstring[temp];
temp++;
}
else
output += process[i];
//output word with no repeated letters
cout << output << endl;
//output asterisks equal to output.length
for (unsigned int a = 0; a < output.length (); a++)
cout << "*";
cout << endl;
//output asterisks for the letter instances removed
for (unsigned int i = 0; i < process.size (); i++)
if (process[i] != '*')
process[i] = ' ';
cout << process << endl << endl;
}
运行代码后我收到的示例输出：
Enter your word below:
INDIA
INDA
****
*
Enter your word below:
abcdefgabchijklmnop
abcdefghijklmnop
****************
***
只需使用简单的数组即可对值进行计数。
#include<iostream>
#include<string>
using namespace std;
int main(){
string s;
char arr[10000];
cin>>s;
int count1[256]={0},count2[256]={0};
for(int i=0;i<s.size();++i){
count1[s[i]]++;
count2[s[i]]++;
}
long max=-1;
int j=0;
for(int i=0;i<s.size();++i){
if(count1[s[i]]==count2[s[i]]){ //check if not printing duplicate
cout<<s[i];
arr[j++]=s[i];
}
if(count2[s[i]]>max)
max=count2[s[i]];
--count1[s[i]];
}
cout<<endl;
for(int i =1; i<=max;++i){
for(int k=0;k<j;++k){
if(count2[arr[k]]){
cout<<"*";
count2[arr[k]]--;
}
else
cout<<" ";
}
cout<<endl;
}
}