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
