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:

  1. Read in and ignore any leading whitespace.
  2. 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.
  3. 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.
  4. Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
  5. 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 than 231 should be clamped to 231, and integers greater than 231 - 1 should be clamped to 231 - 1.
  6. 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:

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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:

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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:

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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:

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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:

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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 <= 200
  • s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+'

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。
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#include <string>
#include <climits>

using namespace std;

class Solution {
public:
    int myAtoi(string s) {
        int n = s.size();
        int i = 0;
        int sign = 1;
        long long result = 0;  // 使用long long暂存结果以检测溢出
        
        // 1. 跳过前置空白字符
        while (i < n && s[i] == ' ') {
            i++;
        }
        
        // 2. 处理正负号
        if (i < n && (s[i] == '+' || s[i] == '-')) {
            sign = (s[i] == '-') ? -1 : 1;
            i++;
        }
        
        // 3. 提取数字并转换
        while (i < n && isdigit(s[i])) {
            int digit = s[i] - '0';
            
            // 检查是否即将溢出
            if (result > INT_MAX / 10 || (result == INT_MAX / 10 && digit > INT_MAX % 10)) {
                return (sign == 1) ? INT_MAX : INT_MIN;
            }
            
            result = result * 10 + digit;
            i++;
        }
        
        // 4. 返回最终结果
        return sign * result;
    }
}

代码解析

  1. 跳过空白字符：使用循环跳过字符串开头的所有空格
  2. 处理符号：
    • 遇到 '+' 保持正号
    • 遇到 '-' 设置为负号
    • 没有符号时默认为正号
  3. 数字转换：
    • 只处理连续的数字字符
    • 每次迭代将当前结果乘以 10 并加上新数字
    • 使用isdigit()函数检查是否为数字字符
  4. 溢出处理：
    • 使用long long类型暂存结果以便检测溢出
    • 当结果即将超过INT_MAX或小于INT_MIN时，返回相应的边界值
    • INT_MAX为 2147483647，INT_MIN为 - 2147483648