leetcode解题: Restore IP Address (93)

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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1： Backtracking, Time O(2^N), Space O(2^N)

也是比较经典的backtracking的题目，也是不停的去取一部分string，要判断不用继续搜索的条件有这些：

  1. 每个字串是否符合ip的条件，一定要是0到255， 并且开头的不能是0， 除了0本身。
  2. 字串的长度不能超过3
  3. ip一共有4部分组成，所以也要判断是否超i过了这个条件。

C++

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class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        if (s.empty()) {
            return res;
        }
        vector<string> cur;
        helper(s, 0, cur, res);
        return res;
    }
    bool isValid(string s) {
        if (s[0] == '0' && s.size() > 1) {
            return false;
        }
        int temp = stoll(s);
        if (temp >= 0 && temp <= 255) {
            return true;
        } else {
            return false;
        }
    }
    string createIP(vector<string>& cur) {
        string res = "";
        for (int i = 0; i < cur.size() - 1; ++i) {
            res += cur[i] + ".";
        }
        res += cur[cur.size() - 1];
        return res;
    }
    void helper(string s, int pos, vector<string>& cur, vector<string>& res) {
        if (pos == s.size()) {
            if (cur.size() == 4) {
                res.push_back(createIP(cur));
            }
            return;
        }
        if (cur.size() > 4) {
            return;
        }
        for (int i = pos; i < s.size(); ++i) {
            if (cur.size() < 4 && i - pos < 3) {
                string item = s.substr(pos, i - pos + 1);
                // check if the item is between 0 and 255
                if (!isValid(item)) {
                    continue;
                }
                cur.push_back(item);
                helper(s, i + 1, cur, res);
                cur.pop_back();
            }
        }
    }
};

Java

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class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        if (s == null || s.length() == 0) {
            return res;
        }
        List<String> current = new ArrayList<String>();
        helper(s, 0, current, res);
        return res;
    }
    boolean isValid(String s) {
        if (s.charAt(0) == '0' && s.length() > 1) {
            return false;
        }
        int temp = Integer.parseInt(s);
        if (temp >= 0 && temp <= 255) {
            return true;
        }
        return false;
    }
    String createIP(List<String> current) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < current.size() - 1; i++) {
            sb.append(current.get(i));
            sb.append(".");
        }
        sb.append(current.get(current.size() - 1));
        return sb.toString();
    }
    private void helper(String s, int pos, List<String> current, List<String> res) {
        if (pos == s.length()) {
            if (current.size() == 4) {
                res.add(createIP(current));
            }
            return;
        }
        if (current.size() > 4) return;
        for (int i = pos; i < s.length() && i < pos + 3; i++) {
            if (current.size() < 4) {
                String item = s.substring(pos, i + 1);
                if (!isValid(item)) {
                    continue;
                }
                current.add(item);
                helper(s, i + 1, current, res);
                current.remove(current.size() - 1);
            }
        }
    }
}