752. IP to CIDR
Difficulty: Easy
Topics: Bit Manipulation
Similar Questions:
Problem:
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.ip + x
(for x < n
) will be a valid IPv4 address.n
will be an integer in the range [1, 1000]
.Solutions:
class Solution {
public:
vector<string> ipToCIDR(string ip, int n) {
vector<string> ret;
size_t start = ipToUnsignedInt(ip);
while (n > 0) {
long longIp = start;
int mask = max(33 - bitLen(longIp & (-longIp)), 33 - bitLen(n));
ret.push_back(unsignedIntToIp(start) + "/" + to_string(mask));
start += (1 << (32 - mask));
n -= (1 << (32 - mask));
}
return ret;
}
private:
size_t ipToUnsignedInt(const string& ip) {
size_t ret = 0;
size_t val= 0;
for (auto& c : ip) {
if (c == '.') {
ret = 256 * ret + val;
val = 0;
} else {
val = 10 * val + c - '0';
}
}
ret = 256 * ret + val;
return ret;
}
string unsignedIntToIp(size_t ip) {
return to_string((ip >> 24) % 256) + "." + to_string((ip >> 16) % 256) + "." + to_string((ip >> 8) % 256) + "." + to_string(ip % 256);
}
size_t bitLen(size_t n) {
if (n == 0) return 1;
size_t ret = 0;
while (n != 0) {
++ret;
n >>= 1;
}
return ret;
}
};