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.

atoi是将字符串转化为整形数的一个函数：char *str = "12345.67"; n = atoi(str); n=12345。

提交过程中遇到几个问题，一个是是否溢出，还有个位数问题，int类型范围（-2147483648~2147483647）超过这个范围就取这两个边界值

int myAtoi(char* str) {
    long long ss=0;
    int i = 0,sign = 1;  //表示符号
    while(str[i]!='' && str[i]==' ') i++;//开头空格舍弃
    if(str[i]=='-') {sign = -1;i++;}
    else if(str[i]=='+') i++;
    int j = i; //****处用到
    for(; str[i] != '';i++){
        if(i > 10 +j && sign == 1) ss = 2147483647;  
        if(i > 10 +j && sign == -1) ss = 2147483648;  //****************************这两行用来判断位数是否超过会溢出的位数 
        if(str[i]>='0' && str[i]<='9') ss = ss*10 + (str[i] - '0');
        if(str[i] < '0' || str[i] > '9') break;
    }
    ss = ss * sign;
    if(ss < -2147483648) return -2147483648;  //判断是否溢出
    if(ss > 2147483647) return 2147483647;
    return ss;
}

如果没有***********处，就会出现如下：

Input:"9223372036854775809"

Output:-2147483648

Expected:2147483647

所以加上了*****************************处的代码

运行了很多次，提交的过程中总会出现各种错误：

2017-10-29 21:27:29