HJ33.整数与IP地址间的转换

左手的ㄟ右手 2024-03-27 14:10 63阅读 0赞

> [牛客华为机试题库【题号 HJ开头】（重点看）][HJ]  
> [牛客在线编程算法篇【题号NC开头】][NC]  
> [剑指offer【题号 JZ开头】][offer_ JZ]  
> [力扣][Link 1]

### 1）[原题链接][Link 2] ###

### 2）已有题解 ###

### 3）代码 ###

package binary;
    
    import java.io.BufferedReader;
    import java.io.IOException;
    import java.io.InputStreamReader;
    import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.List;
    import java.util.stream.Collectors;
    
    import static java.lang.System.in;
    
    public class TranslationBetweenAnIntegerAndAnIpAddress {
        
        public static void main(String[] args) throws IOException {
        
            // 收集数据源
            BufferedReader bf = new BufferedReader(new InputStreamReader(in));
            String temp = "";
            List<String> lines = new ArrayList<>();
            while ((temp = bf.readLine()) != null && temp.length() > 0) {
        
                lines.add(temp);
            }
            bf.close();
    
            // 结果收集容器
            List<Object> res = new ArrayList<>();
    
            // 遍历处理每个IP
            for (int i = 0; i < lines.size(); i++) {
        
                String str = lines.get(i);
                if (str.contains(".")) {
         // 10.0.3.193格式的IP
                    List<Integer> collect = Arrays.stream(str.split("\\.")).map(Integer::parseInt).collect(Collectors.toList());
                    String binaryStr = collect.stream().map(a -> fillProZero(Integer.toBinaryString(a), 8)).map(String::valueOf).collect(Collectors.joining(""));
                    Long decimalIp = binaryToDecimal(binaryStr);
                    res.add(decimalIp);
                } else {
         // 167969729格式的IP
                    String binaryStr = fillProZero(Long.toBinaryString(Long.parseLong(str)), 32);
                    List<String> binarySubStr = get8Str(binaryStr);
                    String dotIp = binarySubStr.stream().map(a -> binaryToDecimal(a)).map(String::valueOf).collect(Collectors.joining("."));
                    res.add(dotIp);
                }
            }
    
            // 输出结果
            res.stream().forEach(System.out::println);
        }
    
    
        /**
         * 字符串8位截取子串
         */
        public static List<String> get8Str(String str) {
        
            List<String> res = new ArrayList<>();
            for (int i = 0; i < str.length(); ) {
        
                if (i + 8 < str.length()) {
        
                    String substring = str.substring(i, i + 8);
                    res.add(substring);
                    i = i + 8;
                } else {
        
                    String substring = str.substring(i);
                    res.add(substring);
                    i = i + 8;
                }
            }
            return res;
        }
    
        /**
         * 不够8位补0
         */
        public static String fillProZero(String str, int base) {
        
            if (str.length() >= base) {
        
                return str;
            }
            return fillProZero("0" + str, base);
        }
    
        /**
         * 2进制到10进制
         */
        public static Long binaryToDecimal(String str) {
        
            char[] chars = str.toCharArray();
            Long res = 0L;
            for (int i = 0; i < chars.length; i++) {
        
                int numericValue = Character.getNumericValue(chars[i]);
                res = res * 2 + numericValue;
            }
            return res;
        }
    }

[HJ]: https://www.nowcoder.com/exam/oj/ta?tpId=37
[NC]: https://www.nowcoder.com/exam/oj
[offer_ JZ]: https://www.nowcoder.com/exam/oj/ta?page=1&tpId=13&type=13
[Link 1]: https://leetcode.cn/problemset/all/
[Link 2]: https://www.nowcoder.com/practice/66ca0e28f90c42a196afd78cc9c496ea?tpId=37&tqId=21256&rp=1&ru=/exam/oj/ta&qru=/exam/oj/ta&sourceUrl=/exam/oj/ta?tpId=37&difficulty=undefined&judgeStatus=undefined&tags=&title=