题解 | #删除字符串中出现次数最少的字符#

基础的解法，基本上没有什么设计而言。

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.nextLine();
        char[] charArr = s.toCharArray();
        // 统计字符串里的各字符出现的次数
        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < charArr.length; i++) {
            map.put(charArr[i], map.getOrDefault(charArr[i], 1) + 1);
        }
        // 获取出现次数最少的字符集合
        int min = 99;
        List<Character> minChar = new ArrayList<>();
        for (char c : map.keySet()) {
            int count = map.get(c);
            if (count < min) {
                min = count;
                minChar.clear();
                minChar.add(c);
            } else if (count == min) {
                minChar.add(c);
            }
        }
        // 输出
        for (int i = 0; i < charArr.length; i++) {
            if (!minChar.contains(charArr[i])) {
                System.out.print(charArr[i]);
            }
        }

    }
}

推荐一篇赞最高的代码，思路很不错

import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            String s = scanner.nextLine();
            char[] chars = s.toCharArray();
            //统计每个字母的数量
            HashMap<Character, Integer> map = new HashMap<>();
            for (char aChar : chars) {
                map.put(aChar, (map.getOrDefault(aChar, 0) + 1));
            }
            //找到数量最少的字符数量
            Collection<Integer> values = map.values();
            Integer min = Collections.min(values);

            //用空字符串替换该字母
            for (Character character : map.keySet()) {
                if (map.get(character) == min){
                    s = s.replaceAll(String.valueOf(character), "");
                }
            }
            System.out.println(s);
        }
    }
}