Note:

All letters in hexadecimal (a-f) must be in lowercase.
The hexadecimal string must not contain extra leading 0s. If the number is zero, it is represented by a single zero character ‘0’; otherwise, the first character in the hexadecimal string will not be the zero character.
The given number is guaranteed to fit within the range of a 32-bit signed integer.
You must not use any method provided by the library which converts/formats the number to hex directly.
Example 1:

Input:
26

Output:
“1a”
Example 2:

Input:
-1

Output:
“ffffffff”

Solution: Simulation

if input is negative, add 2^32 to it (e.g. set the highest bit to 1)
while num is non zero, mod it by 16 and prepend the remainder to ans string, then divide num by 16.

Time complexity: O(logn)
Space complexity: O(logn)

// Author: Huahua
constexpr char kHex[] = "0123456789abcdef";

class Solution {
public:
  string toHex(int num) {
    if (num == 0) return "0";
    long t = num < 0 ? (1LL << 32) + num : num;
    string ans;    
    while (t) {
      ans = kHex[t % 16] + ans;
      t /= 16;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 405. Convert a Number to Hexadecimal appeared first on Huahua's Tech Road.

Given a string num representing a decimal integer N, return the Hexspeak representation of N if it is valid, otherwise return "ERROR".

Example 1:

Input: num = "257"
Output: "IOI"
Explanation:  257 is 101 in hexadecimal.

Example 2:

Input: num = "3"
Output: "ERROR"

Constraints:

• 1 <= N <= 10^12
• There are no leading zeros in the given string.
• All answers must be in uppercase letters.

Solution: Simulation

Time complexity: O(logn)
Space complexity: O(logn)

// Author: Huahua
class Solution {
public:
  string toHexspeak(string num) {
    long n = stol(num);
    string ans;
    while (n) {
      int r = n % 16;
      char c;
      if (r == 0) c = 'O';
      else if (r == 1) c = 'I';
      else if (r >= 10 && r <= 15) c = 'A' + r - 10;
      else return "ERROR";
      ans += c;
      n >>= 4;
    }
    reverse(begin(ans), end(ans));
    return ans;
  }
};

The post 花花酱 LeetCode 1271. Hexspeak appeared first on Huahua's Tech Road.

In the following, every capital letter represents some hexadecimal digit from 0 to f.

The red-green-blue color "#AABBCC" can be written as "#ABC" in shorthand.  For example, "#15c" is shorthand for the color "#1155cc".

Now, say the similarity between two colors "#ABCDEF" and "#UVWXYZ" is -(AB - UV)^2 - (CD - WX)^2 - (EF - YZ)^2.

Given the color "#ABCDEF", return a 7 character color that is most similar to #ABCDEF, and has a shorthand (that is, it can be represented as some "#XYZ"

Example 1:
Input: color = "#09f166"
Output: "#11ee66"
Explanation:  
The similarity is -(0x09 - 0x11)^2 -(0xf1 - 0xee)^2 - (0x66 - 0x66)^2 = -64 -9 -0 = -73.
This is the highest among any shorthand color.

Note:

• color is a string of length 7.
• color is a valid RGB color: for i > 0, color[i] is a hexadecimal digit from 0 to f
• Any answer which has the same (highest) similarity as the best answer will be accepted.
• All inputs and outputs should use lowercase letters, and the output is 7 characters.

Solution: Brute Force

R, G, B are independent, find the closest color for each channel separately.

Time complexity: O(3 * 16)

Space complexity: O(1)

// Author: Huahua
// Running time: 7 ms
class Solution {
public:
  string similarRGB(string color) {
    const string hex{"0123456789abcdef"};
    vector<int> rgb(3, 0);
    for (int i = 0; i < 3; ++i)
      rgb[i] = hex.find(color[2 * i + 1]) * 16 + hex.find(color[2 * i + 2]);
    
    string ans(7, '#');    
    for (int i = 0; i < 3; ++i) {
      int best = INT_MAX;
      for (int j = 0; j < 16; ++j) {
        int diff = abs(j * 16 + j - rgb[i]);
        if (diff >= best) continue;
        best = diff;
        ans[2 * i + 1] = ans[2 * i + 2] = hex[j];
      }
    }
    return ans;
  }
};

The post 花花酱 LeetCode 800. Simple RGB Color appeared first on Huahua's Tech Road.

// 3934609  2017-03-13 08:38:54 Accepted  3713  C++0x 290 6428  xxfflower
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;

void put_hex(int byte) {
   cout << hex << uppercase << setfill('0') << setw(2) << byte;
}

void put_str(const string& s) {
  for(char c : s)
    put_hex(c);
}

void put_len(int length) {
  while(length>=128) {
    put_hex(length % 128 + 128);
    length >>= 7;
  }
  put_hex(length);
}

int main(int argc, char const *argv[])
{
  int n;
  cin>>n;
  string s;
  getline(cin, s);

  while(n--) {
    getline(cin, s);
    put_len(s.length());
    put_str(s);
    cout<<endl;
  }
  return 0;
}

The post [ZOJ] 3713: In 7-bit appeared first on Huahua's Tech Road.