每日经典
《哥舒歌》 ——唐朝西部边民（唐）
北斗七星高，哥舒夜带刀。
至今窥牧马，不敢过临洮。
描述
Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).
The algorithm for myAtoi(string s) is as follows:
- Read in and ignore any leading whitespace.
- Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
- Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
- Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
- If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
- Return the integer as the final result.
Note:
- Only the space character ' ' is considered a whitespace character.
- Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.
Example 1:
Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
^
Step 3: "42" ("42" is read in)
^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.
Example 2:
Input: s = " -42"
Output: -42
Explanation:
Step 1: " -42" (leading whitespace is read and ignored)
^
Step 2: " -42" ('-' is read, so the result should be negative)
^
Step 3: " -42" ("42" is read in)
^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.
Example 3:
Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
Note:
- 0 <= s.length <= 200
- s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.
解析
根据题意，实现 myAtoi(string s) 函数，将字符串转换为 32 位有符号整数。myAtoi(string s) 的算法如下：
- 忽略任何前导空格
- 然后检查下一个字符（如果不在字符串末尾）是“-”还是“+”。如果是，请读入此字符确定最终结果是负数还是正数。如果两者都不存在，则假设结果是正数
- 读入下一个字符，直到到达下一个非数字字符或输入的结尾，字符串的其余部分被忽略。
- 将这些数字转换为整数（即“123”-> 123、“0032”-> 32）。如果未读取任何数字，则整数为 0。根据需要更改符号（从步骤 2 开始）。
- 如果整数超出 32 位有符号整数范围 [-231, 231 - 1]，小于 -231 的整数应限制为 -231，大于 231 - 1 的整数应限制为 231 - 1。
- 返回整数作为最终结果。
其实这个题有很多边界的 case ，这是需要注意的，其他只需要按照上面的算法从前到后写出来即可，我觉的这道题 Acceptance 只有 16% ，可能就是很多人按照自己的思路写代码导致有很多边界 case 报错了，因为我就是这样【狗头】。
解答
class Solution(object):
def myAtoi(self, s):
"""
:type s: str
:rtype: int
"""
if not s : return 0
sign = 1
s = s.lstrip()
if not s: return 0
if s[0] == "+" or s[0] == "-":
sign = 1 if s[0] == "+" else -1
s = s[1:]
print(s)
if not s : return 0
result = 0
for i in range(len(s)):
digit = ord(s[i])-ord('0')
if digit<0 or digit>9: break
result = result * 10 + digit
result *= sign
if result>0:
return result if result<=pow(2,31)-1 else pow(2,31)-1
return result if result>-pow(2,31) else -pow(2,31)
运行结果
Runtime: 22 ms, faster than 69.11% of Python online submissions for String to Integer (atoi).
Memory Usage: 13.6 MB, less than 52.93% of Python online submissions for String to Integer (atoi).
原题链接：leetcode.com/problems/st…
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