题目：

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.

分析：此题不难，但要注意几点

1、首先掠过字符串开始的空格 比如说 S =“        123”，这种最后答案就是 123

2、其次注意正负，若排除前面的空格后，第一个出现的字符是‘ - ’ ，那么按负数处理，第一个出现的是‘ + ’ 或者是数字字符则按正数处理

3、再要注意的就是 正负号禁止出现两次，第二次出现则无需遍历下去

4、在数字字符后若出现了其它字符（除了数字0~9的字符），则字符串遍历结束

5、最后要注意的就是边界问题，32位，所以范围是-2147483648~2147483647 ，超出范围输出边界值

代码：

class Solution {
public:
    int myAtoi(string str) {
        
        int temp = 0, res = 0, i = 0;
        
        //去除空格
        for( ; i < str.size() ; i++)
            if(str[i] != ' ') break;
        
     //   //标记正数
      //  if(str[i] >= '0' && str[i] <= '9')
       //     temp = 0 ;
        
        if(str[i] == '-'){
            i ++ ;
            temp = 1 ;
        }
        else
            if(str[i] == '+')
                i ++ ;
        
        
        for( ; i < str.size() ; i++){
            
            if(str[i] < '0' || str[i] > '9')
                break;
            
            if(temp == 0 && (res > INT_MAX / 10 || (res == INT_MAX / 10 && (str[i] - 48) > 7)))
                return 2147483647 ;
            
            if(temp == 1 && (-1 * res < INT_MIN /10 || (-1 * res == INT_MIN / 10 && (str[i] - 48) >= 8)))
                return -2147483648;
            
            if(str[i] > 47 && str[i] < 58)
                res = res * 10 + (str[i] - 48) ;
        }
        
        if(temp == 1) res = - res ;
        
        return res ;
    }
};