問題描述

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"

補充說明：

一個老生長談的問題，給定一個十進制整數(shù)，輸出它的十六進制形式。這里有幾點要求，其一是輸出的字符均為小寫，其二是如果十六進制的高位為0，那么不顯示多余的0，其三是這個數(shù)字是一個32位整形，其四是不能用轉(zhuǎn)換或者格式化的庫。

方案分析

1. ok，很清晰，正數(shù)用除數(shù)取余處理。負數(shù)處理成正數(shù)再去處理。
2. 負數(shù)如何處理？這里說了，最大是32位整數(shù)，那么就加一個32位的最大值就ok了。

python實現(xiàn)

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

1. 處理數(shù)字問題，不用位操作簡直毀天滅地。4bits表示一個數(shù)字，按照2進制處理唄。
2. 別人的代碼。

python實現(xiàn)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:]