LWC 64: 751. IP to CIDR

LWC 64: 751. IP to CIDR

传送门：751. IP to CIDR

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.
• 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：
题解很取巧，简单说说思路，给定初始的IP之后，转换成2进制的形式，接着每次都找二进制串中的最低位1，它表示的就是CIDR的长度。比如00011000，最低位为00001000，因为在while循环结构内，00011000一定保证在范围内，所以可以认为从00011000开始的step范围内，都是CIDR的某一种解。具体看代码吧。这样一来，我们只需要把00001000转为IP即可，接着让00011000＋00001000，继续求解。具体看代码吧！

Java版本：

    public List ipToCIDR(String ip, int range) {
        long x = 0;
        String[] ips = ip.split("\\.");
        for (int i = 0; i < ips.length; ++i) {
            x = Integer.parseInt(ips[i]) + x * 256;
        }

        List ans = new ArrayList<>();
        while (range > 0) {
            long step = x & -x;  // 最后一个1所在的位置
            while (step > range) step /= 2;
            ans.add(longToIP(x, (int)step));
            x += step;
            range -= step;
        }

        return ans;
    }

    String longToIP(long x, int step) {
        int[] ans = new int[4];
        ans[0] = (int) (x & 255); x >>= 8;
        ans[1] = (int) (x & 255); x >>= 8;
        ans[2] = (int) (x & 255); x >>= 8;
        ans[3] = (int) x;
        int len = 33;
        while (step > 0) {
            len --;
            step /= 2;
        }
        return ans[3] + "." + ans[2] + "." + ans[1] + "." + ans[0] + "/" + len;
    }

Python版本：

    def ipToCIDR(self, ip, range):
        """
        :type ip: str
        :type range: int
        :rtype: List[str]
        """
        ips = ip.split(".")
        x = 0
        for num in ips:
            x <<= 8
            x += int(num)

        ans = []
        while (range > 0):
            step = x & -x
            while (step > range): step /= 2
            ans.append(self.longToIP(x, step))
            x += step
            range -= step
        return ans

    def longToIP(self, x, step):
        ans = [0] * 4
        ans[0] = x & 255
        x >>= 8
        ans[1] = x & 255
        x >>= 8
        ans[2] = x & 255
        x >>= 8
        ans[3] = x
        len = 33
        while (step > 0):
            len -= 1
            step /= 2
        return str(ans[3]) + "." + str(ans[2]) + "." + str(ans[1]) + "." + str(ans[0]) + "/" + str(len)