问题描述
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:26
Output:"1a"
Example 2:
Input:-1
Output:"ffffffff"
补充说明：
一个老生长谈的问题，给定一个十进制整数，输出它的十六进制形式。这里有几点要求，其一是输出的字符均为小写，其二是如果十六进制的高位为0，那么不显示多余的0，其三是这个数字是一个32位整形，其四是不能用转换或者格式化的库。
方案分析
- ok，很清晰，正数用除数取余处理。负数处理成正数再去处理。
- 负数如何处理？这里说了，最大是32位整数，那么就加一个32位的最大值就ok了。
python实现
def toHex(self, num):
ret = ''
map = ('0', '1','2','3','4','5','6','7','8','9','a','b','c','d','e','f')
if num == 0:
return '0'
if num < 0:
num += 2**32
while num > 0 :
num, val = divmod(num, 16)
ret += map[val]
return ret[::-1]
方案分析2
- 处理数字问题，不用位操作简直毁天灭地。4bits表示一个数字，按照2进制处理呗。
- 别人的代码。
python实现2
class Solution(object):
def toHex(self, num):
"""
:type num: int
:rtype: str
"""
if num == 0: return '0'
ret = ''
map = ['a', 'b', 'c', 'd', 'e', 'f']
# 32 bit == 4 byte each x char represents 4 bits, half a byte
for i in xrange(8): # at max we have 32 bit integer, so 8 iterations of computing 4 bits in each iteration == 32 bits
cur = num & 0b1111 # get least significant 4 bits, this corresponds to least significant hex char
char = cur if cur < 10 else map[cur - 10] # fetch hex char
ret = str(char) + ret # append hex char to return
num = num >> 4 # erase the 4 bits we just computed for next iteration
pos = 0
while pos < len(ret) and ret[pos] == '0':
pos += 1
return ret[pos:]
