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

　　這道題的主要難點在於對於異常情況的考慮需要周全，字符數字前面出現字母，正負號，空字符，最大範圍情況。都需要考慮。時間復雜度為O(n), 空間復雜度為O(1)。

解決代碼

 1 class Solution(object):
 2     def myAtoi(self, str):
 3         """
 4         :type str: str
 5         :rtype: int
 6         """
 7         str = str.strip()            # 去除前後的空格
 8         if len(str) < 1:             # 如果長度小於1直接返回0
 9             return 0
10         neg_falg = False             # 設置負數標誌位
11         if str[0] == ‘-‘ or str[0]== ‘+‘:   # 判斷第一位是否帶有正負符號
12             neg_falg = True if str[0] == ‘-‘ else False        # 負號時設置標誌量為負，正好時不做改變。
13             str = str[1:]                                      # 並且去除符號
14         res  = 0                      # 最終結果存儲
15         for i in str: 
16             if ord(i) >= 48 and ord(i) <= 57:                      # 判斷該字符是否屬於數字範圍中
17                 res = res*10 + (ord(i)-ord(‘0‘))                # 求出結果
18             else:
19                 if res == 0:                                    # 如果不屬於，則判斷res結果，直接返回
20                     return 0
21                 break                                           # 終止循環
22         if neg_falg:                                # 根據標誌位來判斷是否為負數
23             res = 0- res
24         if res > pow(2, 31)-1 or res < -pow(2,31):          # 判斷是否溢出
25             return pow(2, 31)-1 if res > pow(2, 31)-1 else -pow(2,31)        
26         return res                                   # 返回結果

【LeetCode每天一題】String to Integer (atoi)(字符串轉換成數字)