[Leetcode]-String to Integer (atoi)

mplement 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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

题目：字串转整数
注意：细节部分非常多，需要考虑很多情况
1、int溢出判断
2、正负号，具体情况见main函数中
3、含有空格的情况
结果：4ms

#include <stdlib.h>#include <stdio.h>#include <math.h>#include <limits.h>int myAtoi(char* str) {   //"NULL"  "000234"  "+23" "-34" "+-34"  "    010"    if(NULL == str )                return 0;        int sum  = 0.0;    int f  = 0;    int DIV = INT_MAX/10;    while('\0' != *str)        {                if(*str == '+')        {            str++;            if(*str < '0' || *str > '9') return 0;            else    f = 0;        }        if(*str == '-')        {            str++;            if(*str < '0' || *str > '9') return 0;            else    f = 1;        }        if(*str >= '0' && *str<= '9' )        {            if(sum > DIV && f == 0) return INT_MAX;            if(sum > DIV && f == 1) return -INT_MAX-1;            sum = sum * 10.0;            if(INT_MAX - *str + '0' < sum  && f == 0)       return INT_MAX;            if((INT_MAX - (*str - '0') < sum ) && (f == 1)) return -INT_MAX-1;            sum = sum + *str - '0';            str++;            if(*str == '\0')                    break;            else if(*str < '0' || *str > '9')   break;        }        else        {            if(*str >= 'a' && *str <= 'z' )     break;            str++;        }    }    if(f) sum = -sum;     return sum;}int main(){    char *str = "0034";    int r = myAtoi(str);    printf("myAtoi str0 is : %d\n",r);    char *str1 = "00";    int r1 = myAtoi(str1);    printf("myAtoi str1 is : %d\n",r1);    char *str2 = "+23";    int r2 = myAtoi(str2);    printf("myAtoi str2 is : %d\n",r2);    char *str3 = "-2345";    int r3 = myAtoi(str3);    printf("myAtoi str3 is : %d\n",r3);    char *str4 = "+-23";    int r4 = myAtoi(str4);    printf("myAtoi str4 is : %d\n",r4);//  expected 0    char *str5 = "    010";    int r5 = myAtoi(str5);    printf("myAtoi str5 is : %d\n",r5);    char *str6 = "    +0104";    int r6 = myAtoi(str6);    printf("myAtoi str6 is : %d\n",r6);    char *str7 = "    -0187";    int r7 = myAtoi(str7);    printf("myAtoi str7 is : %d\n",r7);//  expected -187    char *str8 = "    -018a567";    int r8 = myAtoi(str8);    printf("myAtoi str8 is : %d\n",r8); //  expected -18    char *str9 = "2147483648";    int r9 = myAtoi(str9);    printf("myAtoi str9 is : %d\n",r9); //  expected 2147483647  由于越界，只取最大    char *str10 = "-2147483649";    int r10 = myAtoi(str10);    printf("myAtoi str10 is : %d\n",r10); //  expected -2147483648     char *str11 = "- 204";    int r11 = myAtoi(str11);    printf("myAtoi str11 is : %d\n",r11); //  expected 0      char *str12 = "b3424242";    int r12 = myAtoi(str12);    printf("myAtoi str12 is : %d\n",r12); //  expected 0      while(0);}