文章目录

• 剑指 Offer 20. 表示数值的字符串
• 剑指 Offer 67. 把字符串转换成整数

剑指 Offer 20. 表示数值的字符串

打补丁打到绝望

def isNumber(self, s: str) -> bool:
        m = [str(i) for i in range(10)]
        m += ['+', '-', 'e', 'E', '.']
        res = s.strip()
        for cha in res:
            if cha not in m: return False
        for cha in res:
            if cha == '.' or cha == 'e' or cha == 'E': return self.isEeDot(cha, res)
        return self.isInt(s)

    def isInt(self, ss):
        try: int(ss)
        except: return False
        else: return True

    def isEeDot(self, cha, res):
        if len(res) == 0 or len(res) == 1: return False
        if cha in res:
            li = res.split(cha)
        else: return True
        if len(li) != 2: return False # 保证只有一个e/E/.
        if li[0] == '' and cha == 'e': return False
        elif li[0] == '' and cha == 'E': return False
        if cha == 'e' or cha == 'E':
            if self.isInt(li[-1]) and self.isInt(li[0]): return True
            else: return False
            if not self.isInt(li[-1]): return False # e/E后值必须是整数
        elif cha == '.':
            if ' ' in li[-1] or ' ' in li[0]: return False
            if 'e' in li[-1]: return self.isEeDot('e', li[-1])
            elif 'E' in li[-1]: return self.isEeDot('E', li[-1])
            elif '-' in li[-1] or '+' in li[-1]: return False
            else: return self.isInt(li[-1]) or self.isInt("".join(li[0]))

下面是大佬的代码和思路：












大佬，代码：

class Solution:
    def isNumber(self, s: str) -> bool:
        states = [
            { ' ': 0, 's': 1, 'd': 2, '.': 4 }, # 0. start with 'blank'
            { 'd': 2, '.': 4 } ,                # 1. 'sign' before 'e'
            { 'd': 2, '.': 3, 'e': 5, ' ': 8 }, # 2. 'digit' before 'dot'
            { 'd': 3, 'e': 5, ' ': 8 },         # 3. 'digit' after 'dot'
            { 'd': 3 },                         # 4. 'digit' after 'dot' (‘blank’ before 'dot')
            { 's': 6, 'd': 7 },                 # 5. 'e'
            { 'd': 7 },                         # 6. 'sign' after 'e'
            { 'd': 7, ' ': 8 },                 # 7. 'digit' after 'e'
            { ' ': 8 }                          # 8. end with 'blank'
        ]
        p = 0                           # start with state 0
        for c in s:
            if '0' <= c <= '9': t = 'd' # digit
            elif c in "+-": t = 's'     # sign
            elif c in "eE": t = 'e'     # e or E
            elif c in ". ": t = c       # dot, blank
            else: t = '?'               # unknown
            if t not in states[p]: return False
            p = states[p][t]
        return p in (2, 3, 7, 8)

这道题回头再翻回来看吧。

剑指 Offer 67. 把字符串转换成整数

我写的垃圾代码，写不动了也。

def strToInt(self, str: str) -> int:
        comp = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
        comp.extend(['-', '+'])
        cur = 0
        res = []
        symbol = 0
        while cur < len(str):
            while str[cur] == ' ': cur += 1
            if str[cur] not in comp: return 0
            if str[cur] == '+' or str[cur] == '-': symbol += 1
            if symbol > 1: return 0
            res.append(str[cur])
            cur += 1
        return int("".join(res))

大佬代码：

class Solution:
    def strToInt(self, str: str) -> int:
        str = str.strip()                      # 删除首尾空格
        if not str: return 0                   # 字符串为空则直接返回
        res, i, sign = 0, 1, 1
        int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10
        if str[0] == '-': sign = -1            # 保存负号
        elif str[0] != '+': i = 0              # 若无符号位，则需从 i = 0 开始数字拼接
        for c in str[i:]:
            if not '0' <= c <= '9' : break     # 遇到非数字的字符则跳出
            if res > bndry or res == bndry and c > '7': return int_max if sign == 1 else int_min # 数字越界处理
            res = 10 * res + ord(c) - ord('0') # 数字拼接
        return sign * res

明天起来我再自己写一遍。