算法-一组ip得到没有连续的那个最小的IP

Question

一组ip得到没有连续的那个最小的IP
比如
172.30.1.1，172.30.1.2，172.30.2.4，172.30.1.10
要得到172.30.1.2 这个ip到172.30.1.3 没有连续出现，怎么快速得到172.30.1.2
穷举可以得到。
还有不有其他高性能的算法？请教

Answer 1

兄弟，你这个表述我完全看不懂，写清楚一点吧，可能写清楚一点你自己就知道怎么实现这个逻辑了

不过我猜你应该需要这样的效果:

package com.zycj.code.dici.util;

import java.util.Set;
import java.util.TreeSet;

public class IpFilter {
    private Set<Long> ips = new TreeSet<Long>();

    private IpFilter() {

    }

    private IpFilter(String... ips) {
        for (String ip : ips) {
            addIp(ip);
        }
    }

    public void addIp(String ip) {
        ips.add(ipToLong(ip));
    }

    private long ipToLong(String ip) {
        String[] split = ip.split("\.");
        String hex = "";
        for (String nu : split) {
            hex += numberToHex(nu);
        }
        return Long.parseLong(hex, 16);
    }

    private String numberToHex(String nu) {
        String string = Long.toString(Long.parseLong(nu), 16);
        for (int i = string.length(); i < 2; i++) {
            string = "0" + string;
        }
        return string;
    }

    public String getMinNoLianxu() {
        long current = -1;
        for (Long ip_l : ips) {
            if (current == -1 ) {
                current = ip_l;
            } else {
                if(ip_l%256==1 && ip_l-current==3){//1-254
                    current = ip_l;
                }else if(ip_l-current==1){
                    current = ip_l;
                }else{
                    return longToIp(current);
                }
            }
        }
        return null;
    }

    private String longToIp(long current) {
        String hex = Long.toString(current, 16);
        String result = "";
        for(int i=0;i<4;i++){
            String substring = hex.substring(i*2, i*2+2);
            long parseLong = Long.parseLong(substring, 16);
            String r = Long.toString(parseLong, 10);
            result += r+".";
        }
        return result.substring(0, result.length()-1);
    }
}

测试代码:

public static void main(String[] args) {
        IpFilter ipFilter = new IpFilter();
        ipFilter.addIp("192.168.0.253");
        ipFilter.addIp("192.168.0.254");
        ipFilter.addIp("192.168.1.2");
        ipFilter.addIp("192.168.1.3");
        System.out.println(ipFilter.getMinNoLianxu());
    }

打印结果:

192.168.0.254