需求
代码
class Solution:def myAtoi(self, s: str) -> int:s = s.strip() # 删除首尾空格if not s: return 0 # 字符串为空则直接返回res, i, sign = 0, 1, 1int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10if s[0] == '-': sign = -1 # 保存负号elif s[0] != '+': i = 0 # 若无符号位，则需从 i = 0 开始数字拼接for c in s[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
结果
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