Restore IP Addresses

来源：互联网发布：ubuntu 网桥配置编辑：程序博客网时间：2024/06/09 08:06

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)

典型的回溯法的使用 IP的每一位可以是 1，2，3位一共由4组组成每组的大小为0-255 要注意的是数字组成的合法性 001，01这样的是不合法的要去除掉代码如下：

public class Solution {       public  List<String> restoreIpAddresses(String s) {List<String> res = new ArrayList<String>();ipformat(res, "", "", s, 0, 0, 1, 0);ipformat(res, "", "", s, 0, 0, 2, 0);ipformat(res, "", "", s, 0, 0, 3, 0);return res;}public  void ipformat(List<String> res, String tmp, String sip,String s, int index, int count, int slen, int num) {if (index == s.length() || num == 4) {if (tmp.length() - 4 == s.length()) {String ss="";for(int i=0;i<tmp.length()-1;i++){ss+=tmp.charAt(i);}res.add(ss);}return;}if(slen>=2&&count!=0){    if(sip.charAt(0)=='0') return;}if (count == slen - 1) {int flag = 0;if (slen <= 2) {flag = 1;} else {int com = 0;for (int j = 0; j < slen - 1; j++) {com = (sip.charAt(j) - '0') + com * 10;}com = com * 10 + s.charAt(index) - '0';if (com <= 255&&com>0) {flag = 1;}}if (flag == 1) {if (index == s.length() - 1) {ipformat(res, tmp + sip + s.charAt(index) + '.', "", s,index + 1, 0, 1, num + 1);} else {ipformat(res, tmp + sip + s.charAt(index) + '.', "", s,index + 1, 0, 1, num + 1);ipformat(res, tmp + sip + s.charAt(index) + '.', "", s,index + 1, 0, 2, num + 1);ipformat(res, tmp + sip + s.charAt(index) + '.', "", s,index + 1, 0, 3, num + 1);}}} else {ipformat(res, tmp, sip + s.charAt(index), s, index + 1, count + 1,slen, num);}}}

0 0