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 a whitespace character.
  • Assume we are dealing with an environment that 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: str = "42"
Output: 42

Example 2:

Input: str = "   -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: str = "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

Input: str = "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: str = "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer. Thefore INT_MIN (−231) is returned.

Constraints:

  • 0 <= s.length <= 200
  • s consists of English letters (lower-case and upper-case), digits, ' ', '+', '-' and '.'.

字符串转换整数 (atoi)。

这题考察的是字符串转数字。需要注意的几个点是

1. trim掉字符串中间，前后出现过的所有的空格。例子，"     111 23 4 4   "需要处理成"1112344"，跳过所有的空格。

2. 先判断第一个char是不是一个正负号，若是负号记得最后乘以-1。

3. 对于之后的字符串，判断每个char是不是介于0-9之间，若不是，立马退出循环。若是，就正常计算。因为是从左往右扫描，所以计算方式参见20行。

4. 计算过程中如果有任何时候发现res大于Integer.MAX_VALUE也立马退出循环，并根据sign的正负返回Integer.MAX_VALUE或者Integer.MIN_VALUE

时间O(n)

空间O(1)

Java实现

 1 class Solution {
 2     public int myAtoi(String str) {
 3         // corner case
 4         str = str.trim();
 5         if (str == null || str.length() == 0) {
 6             return 0;
 7         }
 8 
 9         // normal case
10         char firstChar = str.charAt(0);
11         int sign = 1;
12         int start = 0;
13         long res = 0;
14         if (firstChar == '+') {
15             sign = 1;
16             start++;
17         } else if (firstChar == '-') {
18             sign = -1;
19             start++;
20         }
21         for (int i = start; i < str.length(); i++) {
22             if (!Character.isDigit(str.charAt(i))) {
23                 return (int) res * sign;
24             }
25             res = res * 10 + str.charAt(i) - '0';
26             if (sign == 1 && res > Integer.MAX_VALUE) {
27                 return Integer.MAX_VALUE;
28             }
29             if (sign == -1 && res > Integer.MAX_VALUE) {
30                 return Integer.MIN_VALUE;
31             }
32         }
33         return (int) res * sign;
34     }
35 }

 

JavaScript实现

 1 /**
 2  * @param {string} str
 3  * @return {number}
 4  */
 5 var myAtoi = function(str) {
 6     // corner case
 7     str = str.trim();
 8     if (str == null || str.length == 0) {
 9         return 0;
10     }
11 
12     // normal case
13     let firstChar = str.charAt(0);
14     let sign = 1;
15     let i = 0;
16     let res = 0;
17     if (firstChar == '+') {
18         sign = 1;
19         i++;
20     } else if (firstChar == '-') {
21         sign = -1;
22         i++;
23     }
24     while (str.charAt(i) >= '0' && str.charAt(i) <= '9') {
25         res = res * 10 + (str.charAt(i) - 0);
26         if (res * sign >= 2147483647) {
27             return 2147483647;
28         } else if (res * sign <= -2147483648) {
29             return -2147483648;
30         }
31         i++;
32     }
33     return res * sign;
34 };

 

LeetCode 题目总结