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8. String to Integer (atoi) 字符串转成整数

发布时间:2025/4/9 编程问答 41 豆豆
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[抄题]:

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.

 [暴力解法]:

时间分析:

空间分析:

 [优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

[思维问题]:

根本不知道应该怎么处理越界啊:

先设置一个bound变量,-2147483648/10。当前num > bound || num == bond & digit > 7都不行

 

[英文数据结构或算法,为什么不用别的数据结构或算法]:

[一句话思路]:

写整齐点,要考虑到的问题:空格(用trim)、符号(用标记变量)、越界

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

  • 注意要把string转成字符才能操作
  • 字符不是统一处理的 在符号处理和越界处理之后,都要再分别进行i++
  • [二刷]:

  • 整数的范围是 ‘0’  <= c <= '9',必须有等号
  • [三刷]:

  •  num的进位方式是 num = num * 10 + digit digit是最后一位数,不用新相乘
  • [四刷]:

    [五刷]:

      [五分钟肉眼debug的结果]:

    [总结]:

     digit是最后一位数,不用新相乘

    [复杂度]:Time complexity: O(n) Space complexity: O(1)

    [算法思想:迭代/递归/分治/贪心]:

    [关键模板化代码]:

    [其他解法]:

    [Follow Up]:

    [LC给出的题目变变变]:

     [代码风格] :

     [是否头一次写此类driver funcion的代码] :

     [潜台词] :

     

    class Solution {public int myAtoi(String str) {//handle spacestr = str.trim();int i = 0;char[] c = str.toCharArray();//signsint sign = 1;if (i < c.length && (c[i] == '+' || c[i] == '-')) {if (c[i] == '-') sign = -1;i++;}//out of bound in two waysint bound = Integer.MAX_VALUE / 10;int num = 0;while (i < c.length && (c[i] >= '0' && c[i] <= '9')){int digit = c[i] - '0';//out of boundif (num > bound || (num == bound && digit > 7)) {//depend on signreturn (sign == 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE;}num = digit + num * 10;i++;}return num * sign;} } View Code

     

    转载于:https://www.cnblogs.com/immiao0319/p/9383582.html

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