给一个由数字组成的字符串。求出其可能恢复为的所有IP地址。
样例 给出字符串 "25525511135",所有可能的IP地址为:
[ "255.255.11.135", "255.255.111.35" ] (顺序无关紧要)
方法一: 直接三种循环暴力求解
public class Solution {
/**
* @param s the IP string
* @return All possible valid IP addresses
*/
public ArrayList<String> restoreIpAddresses(String s) {
ArrayList<String> res = new ArrayList<String>();
int len = s.length();
for(int i = 1; i<4 && i<len-2; i++){
for(int j = i+1; j<i+4 && j<len-1; j++){
for(int k = j+1; k<j+4 && k<len; k++){
String s1 = s.substring(0,i), s2 = s.substring(i,j), s3 = s.substring(j,k), s4 = s.substring(k,len);
if(isValid(s1) && isValid(s2) && isValid(s3) && isValid(s4)){
res.add(s1+"."+s2+"."+s3+"."+s4);
}
}
}
}
return res;
}
public boolean isValid(String s){
if(s.length()>3 || s.length()==0 || (s.charAt(0)=='0' && s.length()>1) || Integer.parseInt(s)>255)
return false;
return true;
}
}
方法二: 深度搜索,回溯
public class Solution {
/**
* @param s the IP string
* @return All possible valid IP addresses
*/
public ArrayList<String> restoreIpAddresses(String s) {
ArrayList<String> result = new ArrayList<String>();
ArrayList<String> list = new ArrayList<String>();
if(s.length() <4 || s.length() > 12)
return result;
helper(result, list, s , 0);
return result;
}
public void helper(ArrayList<String> result, ArrayList<String> list, String s, int start){
if(list.size() == 4){
if(start != s.length())
return;
StringBuffer sb = new StringBuffer();
for(String tmp: list){
sb.append(tmp);
sb.append(".");
}
sb.deleteCharAt(sb.length()-1);
result.add(sb.toString());
return;
}
for(int i=start; i<s.length() && i < start+3; i++){
String tmp = s.substring(start, i+1);
if(isvalid(tmp)){
list.add(tmp);
helper(result, list, s, i+1);
list.remove(list.size()-1);
}
}
}
private boolean isvalid(String s){
if(s.charAt(0) == '0')
return s.equals("0"); // to eliminate cases like "00", "10"
int digit = Integer.valueOf(s);
return digit >= 0 && digit <= 255;
}
}