# *751. IP to CIDR

Leetcode: 751. IP to CIDR
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:
``````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,

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

x&-x 操作的含义: 取出x中最低为1的那一位，例如:

• 2：二进制为0000 0010，最低为1的那一位是第2位，所以取出后为0000 0010
• 3：二进制为0000 0011，最低为1的那一位是第1位，所以取出后为0000 0001
• 6：二进制为0000 0110，最低为1的那一位是第2位，所以取出后为0000 0010
• 31：二进制为0001 1111，最低为1的那一位是第1位，所以取出后为0000 0001
• 32：二进制为0010 0000，最低为1的那一位是第6位，所以取出后为0010 0000
###### Solution:
``````class Solution {
public  List<String> ipToCIDR(java.lang.String ip, int range) {
long x = 0;
//获得一个ip地址每一部分
String[] ips = ip.split("\\.");
//将整ip地址看为一个整体，求出整体的int表示
for (int i = 0; i < ips.length; ++i) {
x = Integer.parseInt(ips[i]) + x * 256;
}
List<String> ans = new ArrayList<>();
while (range > 0) {
//求出二进制表示下的最低有效位的位数能表示的地址的数量
//如果为奇数，则=1，即以原单个起始ip地址为第一块
//如果为偶数，则二进制表示下的最低有效位的位数能表示的地址的数量
long step = x & -x;
//如果大于range，则需要缩小范围
while (step > range) step /= 2;
//不大于需要的range，开始处理
//求出现在能表示的step个地址的地址块
//x加上以求出的地址块
x += step;
//range减去以表示的地址块
range -= step;
}//直到range<0
return ans;
}
static String longToIP(long x, int step) {
int[] ans = new int[4];
//&255操作求出后8位十进制表示
ans[0] = (int) (x & 255);
//右移8位，即求下一个块
x >>= 8;
ans[1] = (int) (x & 255);
x >>= 8;
ans[2] = (int) (x & 255);
x >>= 8;
ans[3] = (int) x;
int len = 33;
//每一位就可以表示2个
while (step > 0) {
len --;
step /= 2;
}
return ans[3] + "." + ans[2] + "." + ans[1] + "." + ans[0] + "/" + len;
}
}
``````

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