leetcode Restore IP Addresses_coderestore-优快云博客

本文深入探讨了如何通过编程实现一个字符串到有效IP地址的转换过程，并且详细阐述了一个错误的代码实例，以及如何正确修正这个问题。通过逐步分析和代码优化，最终提供了有效的解决方案，帮助开发者理解字符串到特定格式数据的转换过程。

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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)

Have you been asked this question in an interview?

I made a ridiculous mistake in this problem. I once wrote the str2num function like:

  bool str2num(const string& s, int start, int end) {
    int num = 0;
    if (s[start] == '0' && start != end)
      return false;
    
    while (start <= end) {
      //num *= 10;
      num += (s[start] - '0') + num*10;
      //num = num + (s[start] - '0') + num*10;
      ++start;  
    }     
    if (num >= 0 && num <= 255)
      return true;
    return false;
  }

This line seems right, however, this bug torments me all the afternoon. Pls remember, in such case num*=, +=, num never appears in the right side.

num += (s[start] - '0') + num*10;

The following is the right code :

class Solution {
 public:
  vector<string> res;
  int str2num(const string& s, int start, int end) {
    int num = 0;
    while (start <= end) {
      num += (s[start] - '0') + num *10;
      ++start;  
    }     
    return num;
  }
  void restore(const string& s, int start, int depth, vector<string>& cur_str) {
    int num = 0;
    string ip="";
    int left = s.length() - start;
    if (left < 4 - depth || left > (4 - depth) * 3)
      return;
    if (depth == 3 && s.length() - 3 <= start) {
      num = str2num(s, start, s.length() - 1);
      if (num >= 0 && num <= 255) {
        for (int i = 0; i < cur_str.size(); ++i) {
          ip += cur_str[i];
        }
        ip += s.substr(start, s.length() - start);
        res.push_back(ip);
      }
      return;
    }
    else if(depth == 3 && s.length() - 3 > start){
      return;
    }
    else {
      for (int i = 0; i <= 2 && start + i < s.length(); i++ ) {
        num = str2num(s, start, start + i);
        if (num >= 0 && num <= 255) {
          cur_str.push_back(s.substr(start, i+1));
          cur_str.push_back(".");
          restore(s, start + i +1, depth + 1, cur_str);
          cur_str.pop_back();
          cur_str.pop_back();
        }
      }
    }
  }
  vector<string> restoreIpAddresses(string s) {
    // Note: The Solution object is instantiated only once and is reused by each test case.
    res.clear();
    vector<string> cur_str;
    restore(s, 0, 0, cur_str);
    return res;
  }
};

Python Version:

class Solution:
    def isValid(self, s):
        cur_len = len(s)
        if (s[0] == '0' and cur_len > 1):
            return False
        for ch in s:
            if (ch < '0' or ch > '9'):
                return False
        num = int(s)
        return num >= 0 and num <= 255

    def restoreIpAddresses(self, s):
        res = []
        self.dfs(s, [], 4, res)
        return res

    def dfs(self, s, segs, depth, res):
        if (depth == 1):
            if (self.isValid(s)):
                res.append(".".join(segs+[s]))
            return
        cur_len = len(s)
        if (cur_len <= 0):
            return
        for i in range(1,4):
            if (self.isValid(s[0:i]) and i+(depth-1)*1<=cur_len and i+(depth-1)*3>=cur_len):
                self.dfs(s[i:], segs+[s[0:i]], depth-1, res)

s = Solution()
res = s.restoreIpAddresses("")
print(res)