Leetcode 3. Restore IP Addresses
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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)
解法一:
1、IP地址的区域是从0到255
2、建立四个字符串并分别验证它们是否合理
public class Solution { public List<String> restoreIpAddresses(String s) { List<String> res = new ArrayList<>(); int len = s.length(); for(int i = 1; i < 4 ; i++){ for(int j = i+1; j < i+4; j++){ for(int k = j+1; k < j+4 && k < len; k++){ String s1 = s.substring(0,i); String s2 = s.substring(i,j); String s3 = s.substring(j,k); String s4 = s.substring(k,len); if(isValid(s1) && isValid(s2) && isValid(s3) && isValid(s4)){ res.add(s1+"."+s2+"."+s3+"."+s4); } } } } return res; } private 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; }}
解法二:
public class Solution { public List<String> restoreIpAddresses(String s) { List<String> res = new ArrayList<String>(); dfs(s, res, 0, "", 0); return res; } private void dfs(String ip, List<String> res, int len, String temp, int count) { if (count > 4) return; if (count == 4 && len == ip.length()) res.add(temp); for (int i = 1; i < 4; i++) { if (len + i > ip.length()) break; String s = ip.substring(len, len + i); if ((s.startsWith("0") && s.length()>1) || (i==3 && Integer.parseInt(s) >= 256)) continue; dfs(ip, res, len+i, temp + s + (count == 3 ? "" : "."), count + 1); } }
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