解题思路

这是一个字符去重问题,需要保持字符的原始顺序。

关键点:

  1. 使用集合记录已出现的字符
  2. 保持字符的原始顺序
  3. 处理多组输入
  4. 区分大小写

算法步骤:

  1. 读取输入字符串
  2. 使用集合去重
  3. 按原顺序输出

代码

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    string removeDuplicates(const string& str) {
        string result;
        unordered_set<char> seen;
        
        for (char c : str) {
            if (seen.find(c) == seen.end()) {
                result += c;
                seen.insert(c);
            }
        }
        
        return result;
    }
};

int main() {
    string line;
    Solution solution;
    
    while (getline(cin, line)) {
        cout << solution.removeDuplicates(line) << endl;
    }
    
    return 0;
}
import java.util.*;

public class Main {
    static class Solution {
        public String removeDuplicates(String str) {
            StringBuilder result = new StringBuilder();
            Set<Character> seen = new HashSet<>();
            
            for (char c : str.toCharArray()) {
                if (!seen.contains(c)) {
                    result.append(c);
                    seen.add(c);
                }
            }
            
            return result.toString();
        }
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        Solution solution = new Solution();
        
        while (sc.hasNextLine()) {
            String line = sc.nextLine();
            System.out.println(solution.removeDuplicates(line));
        }
        
        sc.close();
    }
}
class Solution:
    def remove_duplicates(self, s: str) -> str:
        seen = set()
        result = []
        
        for c in s:
            if c not in seen:
                result.append(c)
                seen.add(c)
                
        return ''.join(result)

if __name__ == "__main__":
    solution = Solution()
    
    while True:
        try:
            line = input().strip()
            print(solution.remove_duplicates(line))
        except EOFError:
            break

算法及复杂度

  • 算法:集合去重
  • 时间复杂度:,其中 是字符串长度
  • 空间复杂度:,其中 是不重复字符的数量