[Backtracking] [leetcode] Restore the IP address based on the numeric string

topic:

Given a string containing only numbers to represent an IP address, return all valid IP addresses that may be obtained from s. You can return the answers in any order.

A valid IP address consists of exactly four integers (each integer is between 0 and 255, and cannot contain leading 0), and the integers are separated by'.'.

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.

Example 1:

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

source:

93. Recover IP address

Problem-solving ideas: backtracking

Define a variable n, store the value of [start, i], when n meets the IP condition, start recursion + backtracking; if not, exit the loop.

• Recursive termination condition: when the number reaches 4 (start from 0 and reach 4, there is no need to judge the situation greater than 4)
• The result meets the condition: when the number reaches 4, and the character string processing is completed
• Pruning condition: the composition number is greater than 255 (illegal IP number)
• Special treatment: 0 in the IP segment is treated separately

class Solution {
public:
    vector<string> result;
    vector<string> path;
    vector<string> restoreIpAddresses(string s) {
        back(s, 0);
        return result;
    }
    void back(const string& s, int start) {
        if (path.size() == 4) {
            if (start == s.size()) {
                result.push_back(path[0]+"."+path[1]+"."+path[2]+"."+path[3]);
            }
            return;
        }

        if (s[start] == '0') {
            path.push_back("0");
            back(s, start + 1);
            path.pop_back();
            return;
        }
        int n = 0;
        for (int i = start; i < s.size(); i++) {
            n = n * 10 + s[i] - '0';
            if (n > 255) break;
            path.push_back(s.substr(start, i - start + 1));
            back(s, i + 1);
            path.pop_back();
        }
    }
};