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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] 转载地址：http://sgcrj.baihongyu.com/