题解 | #删除字符串中出现次数最少的字符#

import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedHashSet;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String string = scanner.next();
        char[] chars = string.toCharArray();
        LinkedHashSet<String> set = new LinkedHashSet<>();
        ArrayList<String> strList = new ArrayList<>();
        for (int i = 0; i < chars.length; i++) {
            strList.add(chars[i] + "");// 转换成list，为最后删除做准备
            set.add(chars[i] + "");// 去重
        }
        ArrayList<Integer> list = new ArrayList<>();// 存放每个元素的个数
        ArrayList<String> listStr = new ArrayList<>();// 存放去重后的元素
        for (String chStr : set) {
            int count = 0;
            for (int i = 0; i < chars.length; i++) {
                if (chStr.equals(chars[i] + "")) {
                    count++;
                }
            }
            list.add(count);
            listStr.add(chStr);
        }
        Integer min = Collections.min(list);// 找到出现次数最少的次数，再找到他的下标
        ArrayList<Integer> listIndex = new ArrayList<>();// 存放次数最少的下标
        for (int i = 0; i < list.size(); i++) {
            if (min.equals(list.get(i))) {
                listIndex.add(i);
            }
        }
        ArrayList<String> strings = new ArrayList<>();// 存放出现次数最少的元素
        for (int i = 0; i < listIndex.size(); i++) {
            String s = listStr.get(listIndex.get(i));
            strings.add(s);
        }
        for (int i = 0; i < strings.size(); i++) {
            for (int j = 0; j < strList.size(); j++) {
                if (strings.get(i).equals(strList.get(j))) {
                    String removeStr = strList.remove(j);// 删除出现次数最少的元素
                }
            }
        }
        // 得到的strList就是结果，转换成字符串输出。
        for (String s : strList) {
            System.out.print(s);
        }
    }
}