[leetcode]8. String to Integer (atoi)字符串转整数

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

题目

字符串转整数

思路

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代码

 1 class Solution {
 2     public int myAtoi(String str) {
 3         str = str.trim(); // handle str = "    "的情况, 此时str.length() = 4, 不会走第一个if语句，从而会造成后续指针的outofbound
 4         if (str == null || str.length() == 0)
 5         return 0;
 6         
 7         char c = str.charAt(0);
 8         int sign = 1, start = 0, len = str.length();
 9         long sum = 0;
10         // take care of sign 
11         if (c == '+') {
12             sign = 1;
13             start++;
14         } else if (c == '-') {
15             sign = -1;
16             start++;
17         }
18         
19         for (int i = start; i < len; i++) {
20             // take care of non numerical digit, like 'w'
21             if (!Character.isDigit(str.charAt(i))){
22                 return (int) sum * sign;
23             }
24             // convert each char to each digit 
25             sum = sum * 10 + str.charAt(i) - '0';
26             // Input: "-91283472332"   Output:-2147483648
27             if (sign == 1 && sum > Integer.MAX_VALUE){
28                 return Integer.MAX_VALUE;
29             }
30                 
31             if (sign == -1 && (-1) * sum < Integer.MIN_VALUE){
32                 return Integer.MIN_VALUE;
33             }       
34         }
35         return (int) sum * sign;
36     }
37 }