【leetcode刷题笔记】Restore IP Addresses

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)

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题解：深度优先搜索。用result存放得到的每个小数点之间的字符串，在递归函数 private void restoreIpRecur(String s,List<String> answer,ArrayList<String> result){ 中每次从s的头取1个，2个和3个（如果能取到，注意边界判断）组成一个数字，如果该数字在0~255之间（且不是00)，就存入result中备用，然后递归的在s剩下的子串里面搜索后面的数字。

递归终止条件：

• result列表中有四个数字，并且s正好变成空串，说明result中存放的4个数字可以组成一组ip地址，把它存入answer中；
• result列表中有大于或者等于4个数字，但s不为空，说明将s分成了4段以上，不符合规则，return。

代码如下：

public class Solution {
    private boolean isValidIp(String ip){
        if(ip.charAt(0) == '0')
            return ip.equals("0");
        int digit = Integer.valueOf(ip);
        return digit >= 0 && digit <= 255;
    }
    private void restoreIpRecur(String s,List<String> answer,ArrayList<String> result){
        //result has more than four numbers but s is not empty, means we sperate s to more than 4 numbers
        if(result.size()>=4 && !s.isEmpty())
            return;
        if(s.equals("")){
            //if we sperate s exactly four valid numbers
            if(result.size()==4){
                //found one IP
                String ip = new String();
                for(String ss:result)
                    ip = ip + ss + ".";
                ip = ip.substring(0,ip.length()-1);
                answer.add(ip);
            }
            else {
                return;
            }
        }
        //get 1,2,3 characters from s's head and put it into result,search what's left in s recursively
        for(int i = 1;i<=3&&i<=s.length();i++){
            String sub = s.substring(0,i);
            if(isValidIp(sub)){
                result.add(sub);
                restoreIpRecur(s.substring(i), answer, result);
                result.remove(result.size()-1);
            }
        }
        return;
    }
    public List<String> restoreIpAddresses(String s) {
        ArrayList<String> answer = new ArrayList<String>();
        ArrayList<String> result = new ArrayList<String>();
        restoreIpRecur(s, answer, result);
        return answer;
        
    }
}