LeetCode OJ - 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.

分析：基本功考查题

//空                0//以空开头//不是一个数        0//负数             //正数//溢出问题class Solution {public:    int atoi(const char *str) {        if(str == '\0') return 0;                //去掉空格        const char *cur = str;        while(*cur == ' ') {            cur++;           }        if(*cur == '\0') return 0;        //正负判断        int flag = 1;        if(*cur == '-') {            flag = 0;            cur++;        } else if(*cur == '+') {            flag = 1;            cur++;        }         //迭代计算        unsigned long long sum = 0;        while(isdigit(cur)) {            sum = sum * 10 + (*cur - '0');            cur++;        }                //决定最后结果        if(flag && sum >= INT_MAX) return INT_MAX;        if(!flag && sum >= (long long)INT_MAX + 1) return INT_MIN;                return flag ? (int)sum : int(-sum);    }        bool isdigit(const char *ch) {        if(*ch >= '0' && *ch <= '9')            return true;        else             return false;    }};

        关键位置在于溢出判断，这里采用unsigned long long 是64位的无符号长整形，VC++不支持long long但是支持unsigned __int64，调试可以使用这个。对于将unsigned short a 转化为signed int，应先将将a变为int型，后改变符号（补码变原码，即宽度优先原则）。判断负数溢出，这里采用强制类型转化，sum >= (long long)INT_MAX + 1.

        若不采用强制类型转化，这里INT_MAX + 1溢出变为INT_MIN, sum肯定大于INT_MIN。