leetcode - 8. String to Integer (atoi)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

题目大意:实现atoi，考虑各种输入。

大致思路，直接看代码吧

class Solution {
public:
int myAtoi(string str) {
int nNum = str.size();
int nTotal = 0;
int temp = 0;
int nSign = '+';
int i = 0;

//用于溢出判断

//int的最大值
int int_max  = (unsigned int)-1 / 2;

//int的最大值除以10
int temp_max = int_max / 10;

//! 去除字符串前面的空格:' ','\t','\n','\r','\v','\f'
while( (i < nNum) && isspace((int)str[i]))
++i;

//! 如果第一个是正负号跳过它
if (str[i] == '+' || str[i] == '-'){
nSign = str[i];
++i;

/*
//! 我觉得应该跳过它后面的空格，直到第一个非空格字符
//! 但题目认为正负号后面如果接空格直接返回
//! 不过int a = -      123;a却能识别为-123，比较尴尬。。
while( (i < nNum) && isspace((int)str[i]))
++i;
*/
}

//! 检查是否为阿拉伯数字0到9,溢出则返回int的最大或最小值
while((i < nNum) && isdigit((int)str[i])){

//nTotal*10+str[i] - '0
temp = (nTotal<<3) + (nTotal<<1) + str[i] - '0';
++i;

//如果nTotal比temp_max大，乘以10肯定溢出; 如果和突然少于0肯定溢出
if (nTotal > temp_max || temp < 0){
if (nSign == '+')
return int_max;
else
return -int_max - 1;
}
else{
nTotal = temp;
}

}

//!根据符号位返回正数或者负数
if (nSign == '+')
return nTotal;
else
return -nTotal;

}
};

结尾，附上microsoft的 itoa实现版本:

http://blog.csdn.NET/u013266102/article/details/58603255