String to Integer (atoi)
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.
题目分析
将字符串转为int
有几点需要注意的是：
- 字符串前后可能存在空格，应该先丢弃空白字符，找到第一个非空白字符。
- 然后判断符号，复数有负号。输出时也要输出负号。
- 若是第一个非空字符不是符号也不是数字，那么直接返回0。
- 若是超过了int型的上限或者是下限。则输出int型的上下限值。-2147483648 或 2147483647。
代码实在是忍不住用用了网上大神写的，思路太好了，没忍住就抬出来壮壮门面了。好像现在都是贴的人家的程序。咦。这个要注意了以后。
class Solution {
public int myAtoi(String str) {
String strTrim = str.trim(); // remove any leading and trailing whitespace
if (strTrim.length() == 0)
return 0;
int index = 0, sign = 1;
long result = 0;//设置为long型，否则超过上下限就会跳变。无法作比较。
if (strTrim.charAt(index) == '+' || strTrim.charAt(index) == '-') {
sign = strTrim.charAt(index) == '+' ? 1 : -1; // get the sign of the number
index++;
}
// convert string to integer without overflow
while (index < strTrim.length()) {
int digit = strTrim.charAt(index) - '0'; // get the first digit
if (digit < 0 || digit > 9)//当出现了非数字字符，则可以停止了
break;
result = result * 10 + digit; // add the digit to result
if (result > Integer.MAX_VALUE) { // detect the overflow
return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
}
index++;
}
return (int) (result * sign);
}
}