[leetcode: Python]405. Convert a Number to Hexadecimal

Title:
Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, two’s complement method is used.

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:26Output:"1a"

Example 2:

Input:-1Output:"ffffffff"

方法一：58ms

class Solution(object):    def toHex(self, num):        """        :type num: int        :rtype: str        """        dict = {10:'a', 11:'b', 12:'c', 13:'d', 14:'e', 15:'f'}        if num == 0:            return '0'        if num < 0:            num += 2**32        s = ''        s1 = []        if num > 0 and num < 10:            return str(num)        elif num >=10 and num < 16:            return dict[num]        while num >= 16:            num, res =divmod(num, 16)            s1.append(res)        s1.append(num)        for i in s1:            if i < 10:                s = str(i) + s            else:                s = dict[i] + s        return s

方法二：42ms

class Solution(object):    def toHex(self, num):        """        :type num: int        :rtype: str        """        # def get_hex(m, num):        #     res = []        #     while num > 0:        #         res.append(MAP[num % 16])        #         num = num / 16        #     res = ''.join(res[::-1])        #     return res        # def reverse(m, r_m, num):        #     res = []        #     for n in num:        #         res.append(m[15 - r_m[n]])        #     final = ['f'] * 8        #     for i, r in enumerate(res[::-1]):        #         final[7-i] = r        #     return final        # def add_one_on_last(m, r_m, num):        #     c = False        #     if num[7] == 'f':        #         num[7] = '0'        #         c = True        #     else:        #         num[7] = m[r_m[num[7]] + 1]        #     copy_num = num        #     if c:        #         for i, n in enumerate(copy_num[-2::-1]):        #             tmp = r_m[n] + 1        #             if tmp == 16:        #                 num[6-i] = '0'        #             else:        #                 num[6-i] = m[r_m[num[6-i]] + 1]        #                 break        #     return num        # if num == 0:        #     return '0'        # MAP = {        #     i: str(i) for i in range(0,10)        # }        # MAP[10] = 'a'        # MAP[11] = 'b'        # MAP[12] = 'c'        # MAP[13] = 'd'        # MAP[14] = 'e'        # MAP[15] = 'f'        # reverse_map = {v: k for k, v in MAP.items()}        # #print(reverse_map)        # tmp_num = num if num > 0 else -num        # res = get_hex(MAP, tmp_num)        # if num < 0:        #     #print(res)        #     res = reverse(MAP, reverse_map, res)        #     #print(res)        #     res = add_one_on_last(MAP, reverse_map, res)        # return ''.join(res)        if num==0: return '0'        mp = '0123456789abcdef'  # like a map        ans = ''        for i in range(8):            n = num % 16       # this means num & 1111b            c = mp[n]          # get the hex char             ans = c + ans            num = num >> 4        return ans.lstrip('0')  #strip leading zeroes

方法三： 35ms

class Solution(object):    re = {}    def toHex(self, num):        """        :type num: int        :rtype: str        """        if 0 <= num < 10:            return str(num)        if 10 <= num < 16:            return chr(ord('a') + num - 10)        if num not in self.re:            if num < 0:                self.re[num] = self.toHex(2**32+num)            else:                res = num % 16                val = self.toHex(num/16)                self.re[num] = val + self.toHex(res)        return self.re[num]