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:
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.
就是将输入的字符串中的数字提取出来转成int格式。
自己写的代码很啰嗦，被讨论区的代码惊呆到了，很简洁。但是讨论区的代码有瑕疵，for循环没必要。
class Solution {
public:
int myAtoi(string str){
long result=0;
int indicator=1;
if(str.empty())
return 0;
int i=str.find_first_not_of(' ');
if(str[i]=='+'||str[i]=='-'){
indicator=(str[i++]=='-')?-1:1;//注意这里，取到str中的正负号后i++。下标向后移动一位。
}
while(str[i]>='0'&&str[i]<='9'){
result = result*10+(str[i++]-'0');//同样的，str的下标i向后加一位
if(result*indicator>=INT_MAX) return INT_MAX;
if(result*indicator<=INT_MIN) return INT_MIN;
}
return result*indicator;
}
};注意：i++返回原来的值，然后i++。++i返回i+1后的值，i不变。