NowCode：HJ9 提取不重复的整数

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题目：提取不重复的整数

描述

输入一个 int 型整数，按照从右向左的阅读顺序，返回一个不含重复数字的新的整数。

保证输入的整数最后一位不是 0 。

数据范围： 1≤𝑛≤10^8

输入描述：

输入一个int型整数

输出描述：

按照从右向左的阅读顺序，返回一个不含重复数字的新的整数

示例

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输入：
9876673

输出：
37689

题解1

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#include <iostream>
#include <string>
using namespace std;

int main() {
    int input = 0;
    cin >> input;
    string str = to_string(input);
    string ret;
    for (int i = str.size() - 1; i >= 0; i--) {
        if (ret.find(str[i]) == std::string::npos) {
            ret += str[i];
        }
    }
    cout << stoi(ret);
}

题解2

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#include<iostream>
#include<set>
using namespace std;

int main() {
    int input;
    cin >> input;
    set<int> cnt;
    int ans = 0;

    while (input > 0) {
        int temp = input % 10;
        if (cnt.count(temp) == 0) {
            ans = ans * 10 + temp;
            cnt.insert(temp);
        }
        input /= 10;
    }

    cout << ans << endl;
    return 0;
}

思路

我的第一感觉是利用取余不断拿出最后一位，后来想到转成字符串处理再转回整数，即题解1

题解2参考了他人解法，用set记录，用取余开路，用除等于循环

此题很好

strinf::find用法

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// string::find
#include <iostream>       // std::cout
#include <string>         // std::string

int main ()
{
  std::string str ("There are two needles in this haystack with needles.");
  std::string str2 ("needle");

  // different member versions of find in the same order as above:
  std::size_t found = str.find(str2);
  if (found!=std::string::npos)
    std::cout << "first 'needle' found at: " << found << '\n';

  found=str.find("needles are small",found+1,6);
  if (found!=std::string::npos)
    std::cout << "second 'needle' found at: " << found << '\n';

  found=str.find("haystack");
  if (found!=std::string::npos)
    std::cout << "'haystack' also found at: " << found << '\n';

  found=str.find('.');
  if (found!=std::string::npos)
    std::cout << "Period found at: " << found << '\n';

  // let's replace the first needle:
  str.replace(str.find(str2),str2.length(),"preposition");
  std::cout << str << '\n';

  return 0;
}
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first 'needle' found at: 14
second 'needle' found at: 44
'haystack' also found at: 30
Period found at: 51
There are two prepositions in this haystack with needles.