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[Java教程]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 };