标签：ide discard rac white 符号 turned for next sid
问题描述：
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:
‘ ‘ is considered as whitespace character.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.
思路：
根据题目的思路来一步一步的做：
首先先把空格都删除，遇到第一个非空格字符，判断是否是符号‘+’或‘-’来确定数字的符号，若第一个非空格字符串不为符号也不为数字那么就返回0。
若为有效字符需要判断数字的大小。
代码：
class Solution { public: int myAtoi(string str) { if(str.empty()) return 0; int SLen = str.size(); int ret = 0; int sign = 1; int i = 0; while(i < SLen && str[i] == ‘ ‘){ i++; } if(str[i] == ‘-‘ || str[i] == ‘+‘){ sign = str[i] == ‘-‘ ? -1 : 1; i++; } while(i < SLen && str[i] - ‘0‘ > -1 && str[i] - ‘0‘ < 10){ if(ret > INT_MAX / 10 || (ret == INT_MAX / 10 && str[i] - ‘0‘ > 7)) return sign == 1 ? INT_MAX : INT_MIN; ret = ret*10 + (str[i] - ‘0‘); i++; } return sign * ret; } };
需要注意的点：
1. str中的为字符串，一定要经过str[i] - ‘0‘之后才可以当作数字操作：
(ret == INT_MAX / 10 && str[i] - ‘0‘ > 7)
后面的一段判断句中又犯过这种粗心的错误。
2. 前一段判断中要确定ret到底与谁相比较，是INT_MAX/10
标签：ide discard rac white 符号 turned for next sid
原文地址：https://www.cnblogs.com/yaoyudadudu/p/9082527.html
