8. 字符串转换整数 (atoi)
8. 字符串转换整数 (atoi)
题目
Implement the myAtoi(string s)
function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi
function).
The algorithm for myAtoi(string s)
is as follows:
- Read in and ignore any leading whitespace.
- Check if the next character (if not already at the end of the string) is
'-'
or'+'
. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present. - Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
- Convert these digits into an integer (i.e.
"123" -> 123
,"0032" -> 32
). If no digits were read, then the integer is0
. Change the sign as necessary (from step 2). - If the integer is out of the 32-bit signed integer range
[-2^31, 2^31 - 1]
, then clamp the integer so that it remains in the range. Specifically, integers less than-2^31
should be clamped to-2^31
, and integers greater than2^31 - 1
should be clamped to2^31 - 1
. - Return the integer as the final result.
Note:
- Only the space character
' '
is considered a whitespace character. - Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.
Example 1:
Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
Step 2: "42" (no characters read because there is neither a '-' nor '+')
Step 3: " 42 " ("42" is read in)
The parsed integer is 42.
Since 42 is in the range [-2^31, 2^31 - 1], the final result is 42.
Example 2:
Input: s = " -42"
Output: -42
Explanation:
Step 1: " -42" (leading whitespace is read and ignored)
Step 2: " - 42" ('-' is read, so the result should be negative)
Step 3: " - 42 " ("42" is read in)
The parsed integer is -42.
Since -42 is in the range [-2^31, 2^31 - 1], the final result is -42.
Example 3:
Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
Step 3: " 4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
The parsed integer is 4193.
Since 4193 is in the range [-2^31, 2^31 - 1], the final result is 4193.
Constraints:
0 <= s.length <= 200
s
consists of English letters (lower-case and upper-case), digits (0-9
),' '
,'+'
,'-'
, and'.'
.
题目大意
请你来实现一个 myAtoi(string s) 函数,使其能将字符串转换成一个 32 位有符号整数(类似 C/C++ 中的 atoi 函数)。
函数 myAtoi(string s) 的算法如下:
- 读入字符串并丢弃无用的前导空格
- 检查下一个字符(假设还未到字符末尾)为正还是负号,读取该字符(如果有)。 确定最终结果是负数还是正数。 如果两者都不存在,则假定结果为正。
- 读入下一个字符,直到到达下一个非数字字符或到达输入的结尾。字符串的其余部分将被忽略。
- 将前面步骤读入的这些数字转换为整数(即,"123" -> 123, "0032" -> 32)。如果没有读入数字,则整数为 0 。必要时更改符号(从步骤 2 开始)。
- 如果整数数超过 32 位有符号整数范围
[−2^31, 2^31 − 1]
,需要截断这个整数,使其保持在这个范围内。具体来说,小于−2^31
的整数应该被固定为−2^31
,大于2^31 − 1
的整数应该被固定为2^31 − 1
。 - 返回整数作为最终结果。
注意:
- 本题中的空白字符只包括空格字符 ' ' 。
- 除前导空格或数字后的其余字符串外,请勿忽略 任何其他字符。
解题思路
题目要求实现类似 C++
中 atoi
函数的功能,这个函数功能是将字符串类型的数字转成 int
类型数字。
- 先去除字符串中的前导空格,并判断记录数字的符号
sign
; - 遍历字符串,如果
s[i]
是整型,则将其转换成数字类型存入res
; - 如果遇到非整型,则立刻结束遍历;
- 将数字乘以
sign
; - 判断是否超过
int
类型的上限[-2^31, 2^31 - 1]
,如果超过上限,需要输出边界,即-2^31
,或者2^31 - 1
。
复杂度分析
- 时间复杂度:
O(n)
,其中n
是字符串的长度;- 对输入字符串进行
trim()
操作,去除字符串开头和结尾的空格,此操作的时间复杂度为O(n)
; - 遍历字符串,在最坏的情况下,可能需要遍历整个字符串,时间复杂度为
O(n)
;
- 对输入字符串进行
- 空间复杂度:
O(1)
,只使用了几个变量(如i
,res
, 和sign
),没有使用额外的数据结构。
代码
/**
* @param {string} s
* @return {number}
*/
var myAtoi = function (s) {
s = s.trim();
if (s.length == 0) return 0;
let i = 0;
let res = 0;
let sign = 1;
if (s[i] === '-') {
i++;
sign = -1;
} else if (s[i] === '+') {
i++;
}
while (i < s.length && !isNaN(parseInt(s[i]))) {
res = 10 * res + parseInt(s[i]);
i++;
}
res *= sign;
res = Math.max(Math.min(2 ** 31 - 1, res), -(2 ** 31));
return res;
};
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