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.

spoilers alert... click to show requirements for atoi.

Requirements for atoi:

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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

主要是考虑各种各样的输入，对不同的输入进行一个相应的处理

还是挺综合的，考虑问题问题需要很全面，很多的小细节，比如说对越界的判断，对空格，符号的判断 ，这些都是需要谨慎思考的问题

总结下来还是  设计代码的时间要远远长于写代码的时间，思考的过程需要统筹，甚至是实现的小细节；否则实现时考虑很容易出现漏洞，最后甚至自己的代码改的都面目全非了！所以，思考很重要！！！！

class Solution {public:int atoi(const char *str) {int length = strlen(str);int i = 0;//int weight = length-1;//权重，表示当前的处理位是权重，个位为0，十位为1，以此类推int result = 0;bool sign = false; //符号记录标志，表示是否输入符号bool entry = false;//是否进去转化标识int inputSign = 1; //表示输入的符号，1表示正号，-1表示负号if (length == 0)return 0;while (i<length){if (str[i] >= '0' && str[i] <= '9'){entry = true;int temp = (int)str[i] - '0';int tempResult =result*10 + temp;if ((INT_MAX / 10 < result) || (INT_MAX % 10 < temp && INT_MAX / 10 == result)){if (inputSign == 1)return INT_MAX;else if (inputSign == -1)return INT_MIN;}result = tempResult;++i;}else{if (str[i] == ' ' && !entry){++i;}else if ((str[i] == '+' || str[i] == '-') && !sign) //第一次输入符号有效{entry = true;if (str[i] == '-')inputSign = -1;++i;sign = true;}else{if (entry){break;}return 0;}}}return result*inputSign;}};