[LeetCode] 93. 复原IP地址

题目链接 : https://leetcode-cn.com/problems/restore-ip-addresses/

题目描述:

给定一个只包含数字的字符串，复原它并返回所有可能的 IP 地址格式。

示例:

输入: "25525511135"
输出: ["255.255.11.135", "255.255.111.35"]

思路:

我们要知道IP的格式,每位是在0~255之间,

注意: 不能出现以0开头的两位以上数字,比如012,08...

思路一:暴力法

我们把所有出现可能都列举出来,看是否满足条件.

思路二:回溯算法

代码:

思路一:

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        n = len(s)
        res = []
		# 判读是否满足ip的条件
        def helper(tmp):
            if not tmp or (tmp[0] == "0" and len(tmp) > 1) or int(tmp) > 255:
                return False
            return True
		# 三个循环,把数字分成四份
        for i in range(3):
            for j in range(i + 1, i + 4):
                for k in range(j + 1, j + 4):
                    if i < n and j < n and k < n:
                        tmp1 = s[:i + 1]
                        tmp2 = s[i + 1:j + 1]
                        tmp3 = s[j + 1:k + 1]
                        tmp4 = s[k + 1:]
                        # print(tmp1, tmp2, tmp3, tmp4)

                        if all(map(helper, [tmp1, tmp2, tmp3, tmp4])):
                            res.append(tmp1 + "." + tmp2 + "." + tmp3 + "." + tmp4)
        return res

java

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        int n = s.length();
        for (int i = 0; i < 3; i++) {
            for (int j = i + 1; j < i + 4; j++) {
                for (int k = j + 1; k < j + 4; k++) {
                    if (i < n && j < n && k < n) {
                        String tmp1 = s.substring(0, i + 1);
                        String tmp2 = s.substring(i + 1, j + 1);
                        String tmp3 = s.substring(j + 1, k + 1);
                        String tmp4 = s.substring(k + 1);
                        if (helper(tmp1) && helper(tmp2) && helper(tmp3) && helper(tmp4))
                            res.add(tmp1 + "." + tmp2 + "." + tmp3 + "." + tmp4);
                    }
                }
            }
        }
        return res;
    }

    private boolean helper(String tmp) {
        if (tmp == null || tmp.length() == 0 || tmp.length() > 3 || (tmp.charAt(0) == '0' && tmp.length() > 1) || Integer.parseInt(tmp) > 255)
            return false;
        return true;
    }
}

思路二:

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        res = []
        n = len(s)

        def backtrack(i, tmp, flag):
            if i == n and flag == 0:
                res.append(tmp[:-1])
                return
            if flag < 0:
                return
            for j in range(i, i + 3):
                if j < n:
                    if i == j and s[j] == "0":
                        backtrack(j + 1, tmp + s[j] + ".", flag - 1)
                        break
                    if 0 < int(s[i:j + 1]) <= 255:
                        backtrack(j + 1, tmp + s[i:j + 1] + ".", flag - 1)

        backtrack(0, "", 4)
        return res
        

java

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        int n = s.length();
        backtrack(0, "", 4, s, res, n);
        return res;
    }

    private void backtrack(int i, String tmp, int flag, String s, List<String> res, int n) {
        if (i == n && flag == 0) {
            res.add(tmp.substring(0, tmp.length() - 1));
            return;
        }
        if (flag < 0) return;
        for (int j = i; j < i + 3; j++) {
            if (j < n) {
                if (i == j && s.charAt(j) == '0') {
                    backtrack(j + 1, tmp + s.charAt(j) + ".", flag - 1, s, res, n);
                    break;
                }
                if (Integer.parseInt(s.substring(i, j + 1)) <= 255)
                    backtrack(j + 1, tmp + s.substring(i, j + 1) + ".", flag - 1, s, res, n);
            }
        }
    }
}