【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
String to Integer (atoi)
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.
/*
* 问题：atoi字符串转数字函数实现
* 要求：先检测第一个非空格字符，然后识别“+”或“-”，将后面的数字字符翻译为数字，终止于非数字字符
* 数值范围：[−2^31, 2^31 − 1] 即[INT_MIN, INT_MAX]
*/
/*
* 分析参考：
I think we only need to handle four cases:
discards all leading whitespaces
sign of the number
overflow
invalid input
*/
#include<climits>
class Solution
{
public:
int myAtoi(string str)
{
int sign = 1; //符号
long long base = 0; //结果
int i = 0; //用于遍历字符串
if(str.empty()) return 0;
while(str[i] == ' ') i++; //跳过空格
if(str[i] == '+' || str[i] == '-')//翻译正负号
{
sign = (str[i] == '-')? -1: 1;
i++;
}
while(str[i] >= '0' && str[i] <= '9') //数字字符时才进入循环
{
int digit = str[i] - '0'; //翻译数字字符
base = 10*base + digit;
//处理可能的溢出情况
if( (sign>0 && base>INT_MAX) || (sign<0 && -base<INT_MIN) )
{
return (sign==1)? INT_MAX:INT_MIN;
}
i++;
}
return sign*base;
}
};