【LeetCode 10-2】 一 数字（十进制）转化为十六进制

pass: 自己的求解方案，一步一步求解出来
该题求解要求：不能使用库中提供的函数，需要一步一步求解

class Solution:
    def toHex(self, num: int) -> str:
        def get_strOut(num16New):
            strOut = ''
            for numSingle in num16New:
                if numSingle == 15:
                    strOut += 'f'
                elif numSingle == 14:
                    strOut += 'e'
                elif numSingle == 13:
                    strOut += 'd'
                elif numSingle == 12:
                    strOut += 'c'
                elif numSingle == 11:
                    strOut += 'b'
                elif numSingle == 10:
                    strOut += 'a'
                else:
                    strOut += str(numSingle) 
            return strOut

        if num >= 0:
            num16 = []
            while num >= 16:
                num16.append(num % 16)
                num = num // 16
            num16.append(num)
            num16New = num16[::-1]
            strOut = get_strOut(num16New)
        if num < 0:
            num1 = -num
            num16 = []
            while num1 >= 16:
                num16.append(num1 % 16)
                num1 = num1 // 16
            num16.append(num1)


            num16_8 = num16[::-1]
            num16_8_1 = [0] * 8
            num16_8_1[8-len(num16_8):] = num16_8
            #  print(num16_8_1)
            if len(num16_8)==8:  # 碰见了转化后已经达到8位了，提前结束
                return get_strOut(num16_8)

            num16_8_new = []
            num16_8_new.append(15)

            for j in range(1, len(num16_8_1) ):
                num16_8_new.append(15-num16_8_1[j])  # 求反码
            #print(num16_8_new)
            num16_8_new[-1] += 1
            for k in range(len(num16_8_new)-1,0,-1):
                if num16_8_new[k] >= 16:
                    num16_8_new[k] -= 16
                    num16_8_new[k-1] += 1
            strOut = get_strOut(num16_8_new)
        return strOut

大佬的解法
自己还是太弱了

        ans = []
        CONV = "0123456789abcdef"
        # 32位2进制数，转换成16进制 -> 4个一组，一共八组
        for _ in range(8):
            ans.append(num%16)  # 按照这个架势好像反码不影响          # -1 % 16 = 15
            num //= 16
            if not num:
                break
        return "".join(CONV[n] for n in ans[::-1])
        # ans[::-1]  表示进制转换需要倒过来  

利用哈希表
有意思的是这个反码的表达，如果是负值

		result = ''
        hexDict = {0:'0',1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9',10:'a',11:'b',12:'c',13:'d',14:'e',15:'f'}
        if num == 0:
            return '0'
        if num < 0:
            num = 16**8 + num
        while num > 0:
            result = hexDict[num % 16] + result
            num //= 16
        return result