# LeetCode.008.String to Integer (atoi) 字符串转换整数

## 题目描述

8 String to Integer (atoi)
https://leetcode-cn.com/problems/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.

## 解题过程

### 常规从前往后扫描处理

private static class SolutionV2020 {
public int myAtoi(String str) {
if (null == str) {
return 0;
}
// 去掉前后空白
str = str.trim();
if (str.length() == 0) {
return 0;
}
char[] chars = str.toCharArray();
int sign = 1;
int start = 0;
// 处理开头的符号位
if (chars[0] == '-' || chars[0] == '+') {
start = 1;
if (chars[0] == '-') {
sign = -1;
}
} else if (!(chars[0] >= '0' && chars[0] <= '9')) {
return 0;
}
long res = 0;
for (int i = start; i < chars.length; i++) {
if (chars[i] >= '0' && chars[i] <= '9') {
long temp = res * 10 + chars[i] - '0';
if (temp * sign > Integer.MAX_VALUE) {
return Integer.MAX_VALUE;
} else if (temp * sign < Integer.MIN_VALUE) {
return Integer.MIN_VALUE;
}
res = (int)temp;
} else {
break;
}
}
return (int)res * sign;
}
}

### Java正则表达式解决字符串转整型atoi

public int myAtoi2(String str) {
// ( *) 匹配0个或多个空格，[+-]? 匹配0个或1个符号位， \d+ 匹配1个或多个数字， .* 匹配剩余后缀字符
Pattern pattern = Pattern.compile("( *)([+-]?\\d+).*");
Matcher matcher = pattern.matcher(str);
// 先用 matches() 对整个串进行完全匹配
if (!matcher.matches()) {
return 0;
}
// 重置匹配器后，使用 find() 进行部分匹配，其中的第2个捕获组就是合法整数对应的字符串
matcher.reset();
if (matcher.find()) {
String validIntStr = matcher.group(2);
try {
return Integer.parseInt(validIntStr);
} catch (NumberFormatException e) {
return validIntStr.startsWith("-") ? Integer.MIN_VALUE : Integer.MAX_VALUE;
}
}
return 0;
}

## GitHub代码

algorithms/leetcode/leetcode/_008_StringToIntegerAtoi.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_008_StringToIntegerAtoi.java

815

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