8. String to Integer (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 charcter 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
[-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than231should be clamped to231, and integers greater than231 - 1should be clamped to231 - 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 [-231, 231 - 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 [-231, 231 - 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 [-231, 231 - 1], the final result is 4193.
Example 4:
Input: s = "words and 987"
Output: 0
Explanation:
Step 1: "words and 987" (no characters read because there is no leading whitespace)
^
Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
^
Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
^
The parsed integer is 0 because no digits were read.
Since 0 is in the range [-231, 231 - 1], the final result is 0.
Example 5:
Input: s = "-91283472332"
Output: -2147483648
Explanation:
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
^
Step 2: "-91283472332" ('-' is read, so the result should be negative)
^
Step 3: "-91283472332" ("91283472332" is read in)
^
The parsed integer is -91283472332.
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
Constraints:
0 <= s.length <= 200sconsists of English letters (lower-case and upper-case), digits (0-9),' ','+'
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 [−231, 231 − 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 −231 的整数应该被固定为 −231 ，大于 231 − 1 的整数应该被固定为 231 − 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
- 这题是简单题。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。数字需要去掉前导0。最后将数字转换成数字类型，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。
代码实现
package leetcode
func myAtoi(s string) int {
maxInt, signAllowed, whitespaceAllowed, sign, digits := int64(2<<30), true, true, 1, []int{}
for _, c := range s {
if c == ' ' && whitespaceAllowed {
continue
}
if signAllowed {
if c == '+' {
signAllowed = false
whitespaceAllowed = false
continue
} else if c == '-' {
sign = -1
signAllowed = false
whitespaceAllowed = false
continue
}
}
if c < '0' || c > '9' {
break
}
whitespaceAllowed, signAllowed = false, false
digits = append(digits, int(c-48))
}
var num, place int64
place, num = 1, 0
lastLeading0Index := -1
for i, d := range digits {
if d == 0 {
lastLeading0Index = i
} else {
break
}
}
if lastLeading0Index > -1 {
digits = digits[lastLeading0Index+1:]
}
var rtnMax int64
if sign > 0 {
rtnMax = maxInt - 1
} else {
rtnMax = maxInt
}
digitsCount := len(digits)
for i := digitsCount - 1; i >= 0; i-- {
num += int64(digits[i]) * place
place *= 10
if digitsCount-i > 10 || num > rtnMax {
return int(int64(sign) * rtnMax)
}
}
num *= int64(sign)
return int(num)
}
测试代码
package leetcode
import (
"fmt"
"testing"
)
type question8 struct {
para8
ans8
}
// para 是参数
// one 代表第一个参数
type para8 struct {
one string
}
// ans 是答案
// one 代表第一个答案
type ans8 struct {
one int
}
func Test_Problem8(t *testing.T) {
qs := []question8{
{
para8{"42"},
ans8{42},
},
{
para8{" -42"},
ans8{-42},
},
{
para8{"4193 with words"},
ans8{4193},
},
{
para8{"words and 987"},
ans8{0},
},
{
para8{"-91283472332"},
ans8{-2147483648},
},
}
fmt.Printf("------------------------Leetcode Problem 8------------------------\n")
for _, q := range qs {
_, p := q.ans8, q.para8
fmt.Printf("【input】:%v 【output】:%v\n", p.one, myAtoi(p.one))
}
fmt.Printf("\n\n\n")
}
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