Implement atoi which converts a string to an integer.The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value. The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function. If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed. If no valid conversion could be performed, a zero value is returned.
Note:
- Only the space character
' 'is considered as whitespace character. - Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.
Example 1: Input: "42" Output: 42
Example 2: Input: " -42" Output: -42 Explanation: The first non-whitespace character is '-', which is the minus sign. Then take as many numerical digits as possible, which gets 42.
Example 3: Input: "4193 with words" Output: 4193 Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4: Input: "words and 987" Output: 0 Explanation: The first non-whitespace character is 'w', which is not a numerical digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5: Input: "-91283472332" Output: -2147483648 Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer. Thefore INT_MIN (−231) is returned.
思路
这道题的主要难点在于对于异常情况的考虑需要周全，字符数字前面出现字母，正负号，空字符，最大范围情况。都需要考虑。时间复杂度为O(n), 空间复杂度为O(1)。
解决代码
1 class Solution(object):
2 def myAtoi(self, str):
3 """
4 :type str: str
5 :rtype: int
6 """
7 str = str.strip() # 去除前后的空格
8 if len(str) < 1: # 如果长度小于1直接返回0
9 return 0
10 neg_falg = False # 设置负数标志位
11 if str[0] == '-' or str[0]== '+': # 判断第一位是否带有正负符号
12 neg_falg = True if str[0] == '-' else False # 负号时设置标志量为负，正好时不做改变。
13 str = str[1:] # 并且去除符号
14 res = 0 # 最终结果存储
15 for i in str:
16 if ord(i) >= 48 and ord(i) <= 57: # 判断该字符是否属于数字范围中
17 res = res*10 + (ord(i)-ord('0')) # 求出结果
18 else:
19 if res == 0: # 如果不属于，则判断res结果，直接返回
20 return 0
21 break # 终止循环
22 if neg_falg: # 根据标志位来判断是否为负数
23 res = 0- res
24 if res > pow(2, 31)-1 or res < -pow(2,31): # 判断是否溢出
25 return pow(2, 31)-1 if res > pow(2, 31)-1 else -pow(2,31)
26 return res # 返回结果