LeetCode-8：String to Integer (atoi)

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本文详细解析了如何实现字符串转整数(atoi)的功能，并提供了具体的代码实现。讨论了不同输入情况下的处理方式，包括忽略前导空格、处理正负号以及遇到非法字符时的行为。

String to Integer (atoi). 字符串转整数 (atoi).

一、题目

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.

题目翻译：

实现 atoi，将字符串转为整数。

二、解题方案

思路：

字符串类的处理一般考查的都是边界条件、特殊情况的处理。结合本地分析，考虑情况如下：

排除首部的空格，从第一个非空字符开始计算
允许数字以正负号(+-)开头
遇到非法字符便停止转换，返回当前已经转换的值，如果开头就是非法字符则返回0
在转换结果溢出时返回特定值，这里是最大/最小整数

代码实现：

class Solution {
public:
    int myAtoi(string str) {
      str = str.trim();
      int result = 0;
      boolean isPos = true;
      for(int i = 0; i < str.length(); i++){
          char c = str.charAt(i);
          if(i==0 && (c=='-'||c=='+')){
              isPos = c=='+'?true:false;
          } else if (c>='0' && c<='9'){
              // 检查溢出情况
              if(result>(Integer.MAX_VALUE - (c - '0'))/10){
                  return isPos? Integer.MAX_VALUE : Integer.MIN_VALUE;
              }
              result *= 10;
              result += c - '0';
          } else {
              return isPos?result:-result;
          }
      }
      return isPos?result:-result;
    }
};