README_EN.md

93. Restore IP Addresses

中文文档

Description

A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 (inclusive) and cannot have leading zeros.

• For example, "0.1.2.201" and "192.168.1.1" are valid IP addresses, but "0.011.255.245", "192.168.1.312" and "[email protected]" are invalid IP addresses.

Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s. You are not allowed to reorder or remove any digits in s. You may return the valid IP addresses in any order.

 

Example 1:

Input: s = "25525511135"
Output: ["255.255.11.135","255.255.111.35"]

Example 2:

Input: s = "0000"
Output: ["0.0.0.0"]

Example 3:

Input: s = "101023"
Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]

 

Constraints:

• 1 <= s.length <= 20
• s consists of digits only.

Solutions

DFS.

Python3

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        def check(s):
            if not (0 <= int(s) <= 255):
                return False
            if s[0] == '0' and len(s) > 1:
                return False
            return True

        def dfs(s, t):
            if len(t) == 4:
                if not s:
                    ans.append('.'.join(t))
                return
            for i in range(1, min(4, len(s) + 1)):
                if check(s[:i]):
                    t.append(s[:i])
                    dfs(s[i:], t)
                    t.pop()

        ans = []
        dfs(s, [])
        return ans

Java

class Solution {
    private List<String> ans;

    public List<String> restoreIpAddresses(String s) {
        ans = new ArrayList<>();
        dfs(s, new ArrayList<>());
        return ans;
    }

    private void dfs(String s, List<String> t) {
        if (t.size() == 4) {
            if ("".equals(s)) {
                ans.add(String.join(".", t));
            }
            return;
        }
        for (int i = 1; i < Math.min(4, s.length() + 1); ++i) {
            String c = s.substring(0, i);
            if (check(c)) {
                t.add(c);
                dfs(s.substring(i), t);
                t.remove(t.size() - 1);
            }
        }
    }

    private boolean check(String s) {
        if ("".equals(s)) {
            return false;
        }
        int num = Integer.parseInt(s);
        if (num > 255) {
            return false;
        }
        if (s.charAt(0) == '0' && s.length() > 1) {
            return false;
        }
        return true;
    }
}

C++

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> ans;
        vector<string> t;
        dfs(s, t, ans);
        return ans;
    }

    void dfs(string s, vector<string>& t, vector<string>& ans) {
        if (t.size() == 4)
        {
            if (s == "")
            {
                string p = "";
                for (auto e : t) p += e + ".";
                p.pop_back();
                ans.push_back(p);
            }
            return;
        }
        for (int i = 1; i < min(4, (int) s.size() + 1); ++i)
        {
            string c = s.substr(0, i);
            if (check(c))
            {
                t.push_back(c);
                dfs(s.substr(i, s.size() - i), t, ans);
                t.pop_back();
            }
        }
    }

    bool check(string s) {
        if (s == "") return false;
        int num = stoi(s);
        if (num > 255) return false;
        if (s[0] == '0' && s.size() > 1) return false;
        return true;
    }
};

Go

func restoreIpAddresses(s string) []string {
	check := func(s string) bool {
		if i, _ := strconv.Atoi(s); i > 255 {
			return false
		}
		if s[0] == '0' && len(s) > 1 {
			return false
		}
		return true
	}
	var ans []string
	var dfs func(s string, t []string)
	dfs = func(s string, t []string) {
		if len(t) == 4 {
			if s == "" {
				ans = append(ans, strings.Join(t, "."))
			}
			return
		}
		for i := 1; i < 4 && i < len(s)+1; i++ {
			if check(s[0:i]) {
				t = append(t, s[0:i])
				dfs(s[i:], t)
				t = t[0 : len(t)-1]
			}
		}
	}
	var t []string
	dfs(s, t)
	return ans
}

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