转换字符串到整数

转换字符串到整数

实现atoi这个函数，将一个字符串转换为整数。如果没有合法的整数，返回0。如果整数超出了32位整数的范围，返回INT_MAX(2147483647)如果是正整数，或者INT_MIN(-2147483648)如果是负整数。

样例

"10" =>10

"-1" => -1

"123123123123123" => 2147483647

"1.0" => 1

标签

字符串处理 基本实现 优步

  1 class Solution {
  2 public:
  3     /**
  4      * @param str: A string
  5      * @return An integer
  6      */
  7     int atoi(string str) {
  8         // write your code here
  9         str = eraseSpace(str);
 10         int size=str.size();
 11
 12         if(size <= 0)
 13             return 0;
 14
 15         if(str[0]<=‘9‘ && str[0]>=‘0‘)  {
 16             int num=str[0]-‘0‘, i, dec=str[0]-‘0‘;
 17             for(i=1; i<size; i++) {
 18                 if(str[i]<=‘9‘ && str[i]>=‘0‘)  {
 19                     dec = num;
 20                     num = num * 10;
 21                     if(isCrossBorder(num, dec))
 22                         return 2147483647;
 23
 24                     dec = num;
 25                     num += str[i]-‘0‘;
 26                     if(isCrossBorder(num, dec))
 27                         return 2147483647;
 28                 }
 29                 else  {
 30                     return num;
 31                 }
 32             }
 33             return num;
 34         }
 35         else if(str[0] == ‘+‘)  {
 36             int num=0, i, dec=0;
 37             for(i=1; i<size; i++) {
 38                 if(str[i]<=‘9‘ && str[i]>=‘0‘)   {
 39                     dec = num;
 40                     num = num * 10;
 41                     if(isCrossBorder(num, dec))
 42                         return 2147483647;
 43
 44                     dec = num;
 45                     num += str[i]-‘0‘;
 46                     if(isCrossBorder(num, dec))
 47                         return 2147483647;
 48                 }
 49                 else {
 50                     return num;
 51                 }
 52             }
 53             return num;
 54         }
 55         else if(str[0] == ‘-‘)  {
 56             int num=0, i, dec=0;
 57             for(i=1; i<size; i++)  {
 58                 if(str[i]<=‘9‘ && str[i]>=‘0‘)  {
 59                     dec = num;
 60                     num = num * 10;
 61                     if(isCrossBorder(num, dec))
 62                         return -2147483648;
 63
 64                     dec = num;
 65                     num += str[i]-‘0‘;
 66                     if(isCrossBorder(num, dec))
 67                         return -2147483648;
 68                 }
 69                 else  {
 70                     return num*-1;
 71                 }
 72             }
 73             return num*-1;
 74         }
 75         else  {
 76             return 0;
 77         }
 78     }
 79
 80     bool isCrossBorder(int num1, int num2)  {
 81         if(num1 < num2)
 82             return true;
 83         return false;
 84     }
 85
 86   string eraseSpace(string str) {
 87         int size=str.size(),i=0,mark=0;
 88         string result;
 89
 90         for(i=0; i<size; i++)   {
 91             if(str[i]==‘ ‘ && mark==0) {
 92                 continue;
 93             }
 94             else if(str[i]!=‘ ‘ && mark==0)   {
 95                 mark = 1;
 96                 result += str[i];
 97             }
 98             else if(str[i]!=‘ ‘ && mark==1)   {
 99                 mark = 1;
100                 result += str[i];
101             }
102             else if(str[i]==‘ ‘ && mark==1) {
103                 break;
104             }
105         }
106         return result;
107     }
108 };