LeetCode 008 String to Integer (atoi)

【题目】

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.

【题意】

        题意实现atoi库函数，将字符串数组转化为整数

【思路】

    需要考虑以下几种情况
    1. 字符串为空, gui
    2. 整数可正可负，其中"+"可有可无，
        +1,+0,-0也是合法的
    3. 整数越界情况如何处理，溢出下边界则返回INT_MIN， 溢出上边界则返回INT_MAX
    4. 数字中间可能出现空格 "1 2"=>12
    5. 字符串只能由空格 + - 数字四种字符组成，且
        （1）+ - 最多出现一次，且只能出现在所有数字之前
        （2）数字至少出现一次
    
    看了下题目的提示，比上面考虑的要简单的多
    1. 过滤掉首尾的空格，遇到+-后开始向后取最长的连续数字序列，转换成整数即可
        " 11"->11
        "  +12abc"->12
    2. 如果没有这样的数字序列，则不作处理。但这种情况返回什么值呢？题目没说。这里就默认返回0吧

【代码】

class Solution {public:    int atoi(const char *str) {        long long result=0;        int sign=0;   //标记正负符号 -1负号，1正号，0还没有确定正负        const char*p=str;        //找到第一个+-或者数字        while(*p!='\0' && *p==' ')p++;        if(*p!='\0'){            if(*p=='+'){sign=1; p++;}            else if(*p=='-'){sign=-1; p++;}            else if(*p>='0'&&*p<='9'){result=*p-'0'; sign=1; p++;}            else return result;        }        //扫描后续的的连续数字序列        while(*p!='\0'&&*p>='0'&&*p<='9'){            result=result*10+(*p-'0');            p++;        }        //返回结果        result*=sign;        if(result>INT_MAX)result=INT_MAX;        else if(result<INT_MIN)result=INT_MIN;        return (int)result;    }};