Middle-题目105:93. Restore IP Addresses
来源:互联网 发布:泛型 java 方法 编辑:程序博客网 时间:2024/05/16 05:12
题目原文:
Given a string containing only digits, restore it by returning all possible valid IP address combinations.
For example:
Given “25525511135”,
return [“255.255.11.135”, “255.255.111.35”]. (Order does not matter)
题目大意:
给出一个数字组成的字符串,求出能组成的合法IP地址的所有组合。
题目分析:
就是深搜,使用回溯法实现。使用参数dots记录加入了几个点,如果加入了3个点且剩下的字符串是合理的IP数字(无前导0且小于256)就加入最终的列表。这里面有一个剪枝处理是,剩下的数字的长度不大于剩下的点数*3(即形如.255.1000的IP一定不合法)。
比较麻烦的是对0的处理,如果两个点之间的字符串以0开始,则必须是0,”02”,“00”……都不合法。
源码:(language:java)
public class Solution { public List<String> restoreIpAddresses(String s) { List<String> ipList = new ArrayList<String>(); backtrack(ipList,"", 0, s); return ipList; } private void backtrack(List<String> ipList, String currentIP, int dots, String rest) { if(dots==3 && isAvailableIP(rest)) { currentIP+=rest; ipList.add(currentIP); return; } else if(rest.length() <= (4-dots)*3) { for(int i =1;i<=Math.min(rest.length()-1, 3);i++) { String temp = rest.substring(0,i); if(isAvailableIP(temp)) { backtrack(ipList,currentIP+temp+".", dots+1, rest.substring(i)); } } } } private boolean isAvailableIP(String num) { if(num.startsWith("0")) { return num.equals("0"); // 02,00... aren't available. } else { return Integer.parseInt(num)<256; } }}
成绩:
3ms,beats 89.21%,众数4ms,33.87%
cmershen的碎碎念:
此题真的是我原创的,而且是提交到discuss的第一道题。还有一个vote呢~~我想是因为那个对字符串长度的剪枝处理加速了吧。
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