题解 | #字符统计#

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String str = in.next();
        //String str = "aaddccdc";//答案cda
        char[] cary = str.toCharArray();
        int[] iary = new int[123]; //ascii码最大为z=122
        for (char i : cary) {
            iary[(int)i]++;//给码值对应的索引累加计数
        }
        int max = 0;
        for (int i : iary) {
            if (max < i) {
                max = i; //获取最大值
            }
        }
        StringBuilder sb = new StringBuilder();
        while (max >
                0) { //对数组一次循环只取当前最大值,此时ascii码已按从小到大排着的
            for (int i = 0; i < iary.length; i++) {
                if (iary[i] == max) {
                    sb.append((char)i);
                }
            }
            max--;
        }
        System.out.println(sb);
    }
}

网上看到的做法,以空间换复杂度,还行吧,

z是122,所有int[123]就是最大的,所有进位什么的都不变,完全不动脑