String to Integer(atoi) LeetCode

题目链接点击此处，题目描述如下：

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.

题目表面看起来比较简单，但是坑不少，什么滤除前缀空格、正负号就不提了。主要问题在于清楚理解数据的表示范围。

__int64 INT_MAX_P_1 = INT_MAX + 1;

并不能得到真正比INT_MAX大1的数字(由于INT_MAX是int，所以加1之后还是int，并没有自动的变为__int64)，如果想加1之后自动变为__int64，需要分开写，即：

__int64 INT_MAX_P_1 = INT_MAX；INT_MAX_P_1 += 1;

再者，需要注意输入为INT_MIN时的处理，即输入为-2147483648时的处理

最后又来了一个坑，尼玛，测试用例竟然有一个65位的数字："18446744073709551617"

思前想后，还是在将字符串转换为数字的时候，顺便记录转换后的长度，如果长度超过11位，同样根据isNative的符号，直接处理，完整代码：

int myAtoi(char* str) {long long int tmpResult = 0;int isNaive = 1;int result = 0;long long int ATOI(const char* str, int *length);int i = 0;int numLength = 0;    long long int INT_MAX_P_1 = INT_MAX;INT_MAX_P_1 += 1;//滤除空格for (i=0; ' ' == str[i]; ++i);if ('-' == str[i]){isNaive = -1;++i;}else if ('+' == str[i]){isNaive = 1;++i;}if (str[i] >= '0' && str[i] <= '9'){tmpResult = ATOI(&str[i], &numLength);}    if (tmpResult < 0 || INT_MAX_P_1 <= tmpResult || numLength >= 11){if (-1 == isNaive){result = INT_MIN;}else if (1 == isNaive){result = INT_MAX;}}else{result = isNaive * tmpResult;}return result;}long long int ATOI(const char* str, int *length){int i = 0;long long int result = 0;while (str[i] >= '0' && str[i] <= '9'){result = result * 10 + (str[i] - '0');++i;}(*length) = i;return result;}