【题目描述】

给你一个混合字符串 s ，请你返回 s 中 第二大 的数字，如果不存在第二大的数字，请你返回 -1 

混合字符串 由小写英文字母和数字组成。

https://leetcode.cn/problems/second-largest-digit-in-a-string/description/

【代码】Leecode-set

package com.company;
// 2023-04-05;
import java.util.*;

class Solution {
    public int secondHighest(String s) {
       Set<Integer> set = new HashSet<>();
       for (char c : s.toCharArray()){
           if (Character.isDigit(c)){
               // 字符转数字了
               set.add(c - '0');
           }
       }
        Object[] obj = set.stream().sorted().toArray();
        return set.size() < 2 ? -1 : (int)obj[set.size() - 2]; 
    }
}

public class Test {
    public static void main(String[] args) {
        new Solution().secondHighest("dfa12321afd"); // 输出：2
        new Solution().secondHighest("abc1111"); // 输出：-1
        new Solution().secondHighest("1a0"); // 输出：0
    }
}

【代码】Leecode-一次遍历

package com.company;
// 2023-04-05;
import java.util.*;

class Solution {
    public int secondHighest(String s) {
        int mv1 = -1;
        int mv2 = -1;
        for (char c : s.toCharArray()){
            int t = c - '0';
            if (c < '0' || c > '9' || t == mv1) continue;
            if (t > mv1) {
                mv2 = mv1;
                mv1 = t;
            }else if ((t > mv2)){
                mv2 = t;
            }
        }
        return mv2;
    }
}

public class Test {
    public static void main(String[] args) {
        new Solution().secondHighest("dfa12321afd"); // 输出：2
        new Solution().secondHighest("abc1111"); // 输出：-1
        new Solution().secondHighest("1a0"); // 输出：0
    }
}


【代码】admin

思路：1. 去除所有的字符, 那么就只剩下数字了

          2. 利用set去重, 排序

          3. 把set转Object数组, 然后返回倒是第2个元素。如果单元素返回-1

package com.company;
// 2023-04-05;
import java.util.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.stream.Collectors;

class Solution {
    public int secondHighest(String s) {
        if (s.length() == 1) return -1;
        int num = 1;
        s = s.replaceAll("[a-zA-Z]+", "");
        Matcher matcher = Pattern.compile("\\d+").matcher(s);
        if (matcher.find()){
            String group = matcher.group();
            char[] chars = group.toCharArray();
            int[] res = new int[chars.length];
            for (int i = 0; i < chars.length; i++){
                res[i] = Character.getNumericValue(chars[i]);
            }
            Set<Integer> collect = Arrays.stream(res).boxed().sorted().collect(Collectors.toSet());
            if (collect.size() == 1){
                return -1;
            }else {
                Object[] objects = collect.toArray();
                num = (int) objects[objects.length - 2];
            }
        }else {
            return -1;
        }
        return num;
    }
}

public class Test {
    public static void main(String[] args) {
        new Solution().secondHighest("dfa12321afd"); // 输出：2
        new Solution().secondHighest("abc1111"); // 输出：-1
        new Solution().secondHighest("1a0"); // 输出：0
    }
}