HJ9 提取不重复的整数 HJ09

​​HJ9 提取不重复的整数​​

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

class HJ09 {
private:
    static vector<int> digits(int num) {
        vector<int> vi;
        int d = 0;
        while (num != 0) {
            d = num % 10;
            vi.push_back(d);
            num /= 10;
        }
        return vi;
    }
    static int distinctDigits(vector<int> v) {
        int a[] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
        int ans = 0;
        for (vector<int>::const_iterator it = v.begin(); it != v.end(); ++it) {
            if (a[*it] == 0) {
                ans = 10 * ans + *it;
                a[*it] += 1;
            }
        }
        return ans;
    }
public:
    int solution(int num) {
        vector<int> v = digits(num);
        return distinctDigits(v);
    }
};
 
int main() {
    HJ09 q;
    int num = 0;
    cin >> num;
    cout << q.solution(num);
}

总共只有0-9 数字，就不需要set了

上一篇: HJ8 合并表记录 HJ08

​​HJ10 字符个数统计​​

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

int main(){
    string line;
    set<char> cs;
    cin >> line;
    
    for (int i = 0; i < line.size(); i++) {
        cs.insert(line.at(i));
    }
    cout << cs.size() << "\n";
}

set用法，以上。

​​HJ11 数字颠倒​​

还是一样的套路，数字转字符串

#include <iostream>
using namespace std;

class HJ11 {
public:
    const char *solution(int num) {
        if (num == 0) {
            return "0";
        }
        int d = 0, i = 0;
        // 数据范围 0<=n && n <= Math.pow(2, 30)-1
        char ans[30] = {'\0'};
        while (num != 0) {
            d = num % 10;
            num /= 10;
            ans[i++] = '0' + d;
        }
        return ans;
    }
};

int main() {
    HJ11 q;
    int num = 0;
    
    cin >> num;
    const char *ans = q.solution(num);

    cout << ans << endl;
}