[Leetcode] 751. IP to CIDR 解题报告

最新推荐文章于 2024-08-11 16:21:57 发布

原创最新推荐文章于 2024-08-11 16:21:57 发布 · 921 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #解题报告 #Bit manipulation

IT公司面试习题专栏收录该内容

736 篇文章

订阅专栏

本文介绍了LeetCode上751题的解题报告，主要涉及IP地址和CIDR块的转换。通过对给定起始IP和要覆盖的IP数量，计算并返回最小长度的CIDR块列表。文章包含题目描述、示例及解题思路，并附有代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

Given a start IP address ip and a number of ips we need to cover n, return a representation of the range as a list (of smallest possible length) of CIDR blocks.

A CIDR block is a string consisting of an IP, followed by a slash, and then the prefix length. For example: "123.45.67.89/20". That prefix length "20" represents the number of common prefix bits in the specified range.

Example 1:

Input: ip = "255.0.0.7", n = 10
Output: ["255.0.0.7/32","255.0.0.8/29","255.0.0.16/32"]
Explanation:
The initial ip address, when converted to binary, looks like this (spaces added for clarity):
255.0.0.7 -> 11111111 00000000 00000000 00000111
The address "255.0.0.7/32" specifies all addresses with a common prefix of 32 bits to the given address,
ie. just this one address.

The address "255.0.0.8/29" specifies all addresses with a common prefix of 29 bits to the given address:
255.0.0.8 -> 11111111 00000000 00000000 00001000
Addresses with common prefix of 29 bits are:
11111111 00000000 00000000 00001000
11111111 00000000 00000000 00001001
11111111 00000000 00000000 00001010
11111111 00000000 00000000 00001011
11111111 00000000 00000000 00001100
11111111 00000000 00000000 00001101
11111111 00000000 00000000 00001110
11111111 00000000 00000000 00001111

The address "255.0.0.16/32" specifies all addresses with a common prefix of 32 bits to the given address,
ie. just 11111111 00000000 00000000 00010000.

In total, the answer specifies the range of 10 ips starting with the address 255.0.0.7 .

There were other representations, such as:
["255.0.0.7/32","255.0.0.8/30", "255.0.0.12/30", "255.0.0.16/32"],
but our answer was the shortest possible.

Also note that a representation beginning with say, "255.0.0.7/30" would be incorrect,
because it includes addresses like 255.0.0.4 = 11111111 00000000 00000000 00000100 
that are outside the specified range.

Note:

ip will be a valid IPv4 address.
Every implied address ip + x (for x < n) will be a valid IPv4 address.
n will be an integer in the range [1, 1000].

思路：

没看到题目是什么意思，先贴个别人的解法，然后仔细琢磨。

代码：

class Solution {
public:
    vector<string> ipToCIDR(string ip, int n) {
        unsigned int IP = 0;
        string temp;
        for(int i = 0; i <= ip.size(); ++i) {
            if(ip[i] == '.' || i == ip.size()) {
                IP |= stoi(temp); 
                if(i != ip.size()) {
                    IP <<= 8;
                }
                temp.clear();
            }
            else temp.append(1, ip[i]);
        } 
        IP += n;   
        vector<string> res;
        while(n > 0) {
            unsigned int temp = IP & (IP - 1);
            int dif = IP - temp;
            while(dif > n) {
                dif /= 2;
            }
            IP -= dif;
            n -= dif; 
            res.push_back(convert(IP, dif));
        }
        reverse(res.begin(), res.end());
        return res;
    }
private:
    string convert(unsigned int a, int dif) {
        string res = "";
        for(int i = 0; i < 4; ++i) {
            int temp = a & 255; 
            res = to_string(temp) + "." + res;
            a >>= 8;
        }
        res.back() = '/';  
        int len = 1;
        while(dif != 1) {
            dif >>= 1;
            ++len;
        }
        return res + to_string(32 - len + 1);
    }
};