LeetCode第八题：String to Integer (atoi)（C++）
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.
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.
题目链接：https://leetcode.com/problems/string-to-integer-atoi/
解题代码：
解题思路：
1.去除字符串中空字符串
if(str[i] == ’ ')
{
continue;
}
2.判断非空字符串首字符 + -字符，设置标志位flag
flag = (str[i++] == ‘+’) ? 1 : -1; //此时i位置为+ 或-；i++，从下一位开始
3.获取整数数字
result = result * 10 + (str[i++] - ‘0’);//i++是进位
解释：减去字符0，也就是减去0的ASCII码值48，数字字符减去‘0’即可得到了该数字
性能：
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