LeetCode 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)

题意：给出一串字符串，求可能的合法IP个数。

思路：dfs处理，最多就4层递归，因为ip最多就是由4个数组成的。

public class Solution {   public List<String> restoreIpAddresses(String s) {    List<String> result = new ArrayList<String>();    if (s.length() < 4 || s.length() > 15) return result;        List<String> tmp = new ArrayList<String>();    dfs(result, tmp, 0, 0, s);    return result;    }        private void dfs(List<String> result, List<String> tmp, int dep, int cur, String s) {if (dep >= 4) {if (cur == s.length()) {StringBuilder ans = new StringBuilder();ans.append(tmp.get(0));for (int i = 1; i < 4; i++)ans.append("." + tmp.get(i));result.add(ans.toString());}return;}for (int k = 1; k <= 3 && cur+(k-1) < s.length(); k++) {String num = s.substring(cur, cur+k);if (isValid(num)) {tmp.add(num);dfs(result, tmp, dep+1, cur+k, s);tmp.remove(tmp.size()-1);}}}private boolean isValid(String num) {if (num.length() == 1 && num.compareTo("0") >= 0 && num.compareTo("9") <= 0) return true;if (num.length() == 2 && num.compareTo("10") >= 0 && num.compareTo("99") <= 0) return true;if (num.length() == 3 && num.compareTo("100") >= 0 && num.compareTo("255") <= 0) return true;return false;}}

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