题解 | #删除字符串中出现次数最少的字符#

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            String a = in.next();
            System.out.println(change(a));
        }
    }

    private static String change(String a) {
        HashMap<String, Integer> map = new HashMap<>();
        //统计每个字符出现的次数
        int length = a.length();
        for (int i = 0; i < length; i++) {
            String s = String.valueOf(a.charAt(i));
            if (map.containsKey(s)) {
                map.put(s, map.get(s) + 1);
            } else {
                map.put(s, 1);
            }
        }
        //获取出现次数最少的字符串集合
        ArrayList<String> list = new ArrayList<>();
        for (Integer i = 1; i < length; i++) {
            if (map.containsValue(i)) {
                Integer finalI = i;
                map.forEach((s1, s2) -> {
                    if (finalI.equals(s2)) {
                        list.add(s1);
                    }
                });
                break;
            }
        }
        //根据集合记录修改字符串
        for (String s : list) {
            a = a.replaceAll(s, "");
        }
        return a;
    }
}

解题思路:

1, 统计每个字符出现的次数;

2, 获取出现次数最少的字符串集合;

3, 根据集合记录修改字符串