LeetCode 751. IP to CIDR

原题链接在这里：https://leetcode.com/problems/ip-to-cidr/

题目：

An IP address is a formatted 32-bit unsigned integer where each group of 8 bits is printed as a decimal number and the dot character '.' splits the groups.

• For example, the binary number 00001111 10001000 11111111 01101011 (spaces added for clarity) formatted as an IP address would be "15.136.255.107".

A CIDR block is a format used to denote a specific set of IP addresses. It is a string consisting of a base IP address, followed by a slash, followed by a prefix length k. The addresses it covers are all the IPs whose first k bits are the same as the base IP address.

• For example, "123.45.67.89/20" is a CIDR block with a prefix length of 20. Any IP address whose binary representation matches 01111011 00101101 0100xxxx xxxxxxxx, where x can be either 0 or 1, is in the set covered by the CIDR block.

You are given a start IP address ip and the number of IP addresses we need to cover n. Your goal is to use as few CIDR blocks as possible to cover all the IP addresses in the inclusive range [ip, ip + n - 1] exactly. No other IP addresses outside of the range should be covered.

Return the shortest list of CIDR blocks that covers the range of IP addresses. If there are multiple answers, return any of them.

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 IP addresses that need to be covered are:
- 255.0.0.7  -> 11111111 00000000 00000000 00000111
- 255.0.0.8  -> 11111111 00000000 00000000 00001000
- 255.0.0.9  -> 11111111 00000000 00000000 00001001
- 255.0.0.10 -> 11111111 00000000 00000000 00001010
- 255.0.0.11 -> 11111111 00000000 00000000 00001011
- 255.0.0.12 -> 11111111 00000000 00000000 00001100
- 255.0.0.13 -> 11111111 00000000 00000000 00001101
- 255.0.0.14 -> 11111111 00000000 00000000 00001110
- 255.0.0.15 -> 11111111 00000000 00000000 00001111
- 255.0.0.16 -> 11111111 00000000 00000000 00010000
The CIDR block "255.0.0.7/32" covers the first address.
The CIDR block "255.0.0.8/29" covers the middle 8 addresses (binary format of 11111111 00000000 00000000 00001xxx).
The CIDR block "255.0.0.16/32" covers the last address.
Note that while the CIDR block "255.0.0.0/28" does cover all the addresses, it also includes addresses outside of the range, so we cannot use it.

Example 2:

Input: ip = "117.145.102.62", n = 8
Output: ["117.145.102.62/31","117.145.102.64/30","117.145.102.68/31"]

Constraints:

• 7 <= ip.length <= 15
• ip is a valid IPv4 on the form "a.b.c.d" where a, b, c, and d are integers in the range [0, 255].
• 1 <= n <= 1000
• Every implied address ip + x (for x < n) will be a valid IPv4 address.

题解：

First convert ip address to long val.

Find the right most 1. That means how many numbers it could cover. e.g. "1000", right most 1 is 8. It covers the next 8 numbers to "1111".

Subtract this number step from n. 

while n is smaller than step, then step /= 2. 

Add this converted CIDR to the res.

Note: 1.if val = 0, there is no right 1. step = 0. Then step = 1. Then step is 2^x, right bigger than n.

2. When convert val and step to CIDR. The first number & 255 is the last part of CIDR.

Time Complexity: O(1). Because it is at most 32 bits.

Space: O(1). regardless res.

AC Java:

class Solution {
    public List<String> ipToCIDR(String ip, int n) {
        long val = 0;
        String [] ips = ip.split("\\.");
        for(int i = 0; i < ips.length; i++){
            val = val * 256 + Integer.valueOf(ips[i]);
        }
        
        List<String> res = new ArrayList<>();
        while(n > 0){
            long step = val & -val;
            if(step == 0){
                step = 1;
                while(step < n){
                    step *= 2;
                }
            }
            while(step > n){
                step /= 2;
            }
            
            res.add(longToIp(val, (int)step));
            n -= step;
            val += step;
        }
        
        return res;
    }
    
    private String longToIp(long val, int step){
        int [] res = new int[4];
        for(int i = 0; i < 4; i++){
            res[i] = (int)(val & 255);
            val >>= 8;
        }
        
        int len = 32;
        while(step > 1){
            len--;
            step /= 2;
        }
        
        return res[3] + "." + res[2] + "." + res[1] + "." + res[0] + "/" + len;
    }
}