LeetCode 8 String to Integer (string转int)

题目来源：https://leetcode.com/problems/string-to-integer-atoi/

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

 

解题思路：

照着要求写代码，可以总结如下：

1. 字串为空或者全是空格，返回0； 

2. 字串的前缀空格需要忽略掉；

3. 忽略掉前缀空格后，遇到的第一个字符，如果是‘+’或‘－’号，继续往后读；如果是数字，则开始处理数字；如果不是前面的2种，返回0；

4. 处理数字的过程中，如果之后的字符非数字，就停止转换，返回当前值；

5. 在上述处理过程中，如果转换出的值超出了int型的范围，就返回int的最大值或最小值。

 

Java代码：

 1 public class Solution {
 2     public int myAtoi(String str) {
 3         int max = Integer.MAX_VALUE;
 4         int min = -Integer.MIN_VALUE;
 5         long result = 0;
 6         str = str.trim();
 7         int len = str.length();
 8         if (len < 1)
 9             return 0;
10         int start = 0;
11         boolean neg = false;
12  
13         if (str.charAt(start) == '-' || str.charAt(start) == '+') {
14             if (str.charAt(start) == '-')
15                 neg = true;
16             start++;
17         }
18  
19         for (int i = start; i < len; i++) {
20             char ch = str.charAt(i);
21  
22             if (ch < '0' || ch > '9')
23                 break;
24             result = 10 * result + (ch - '0');
25             if (!neg && result > max)
26                 return max;
27             if (neg && -result < min)
28                 return min;
29  
30         }
31         if (neg)
32             result = -result;
33  
34         return (int) result;
35     }
36  
37 };

 

转载于:https://www.cnblogs.com/zpfbuaa/p/5076830.html