【leetcode】String to Integer (atoi)

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.

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.

atoi的原理：

1.如果字符串前面有空格，则去除所有空格

2.允许有正负号,需要读取出正负号，连续两个正负号则是错误的字符串

3.读取过正负号后，开始读取数字，直到读到不是数字的位置或者字符串结束时停止，解析出里面的数字

注意会溢出，溢出时返回INT_MAX或者INT_MIN

 1 class Solution {
 2 public:
 3     int atoi(const char *str) {
 4         if(str==NULL) return 0;
 5         int result=0;
 6         int flag=1;
 7         while(*str==‘ ‘)
 8         {
 9             str++;
10         }
11         if(*str==‘+‘)
12         {
13             flag=1;
14             str++;
15         }
16         else if(*str==‘-‘)
17         {
18             flag=-1;
19             str++;
20         }
21         while(*str>=‘0‘&&*str<=‘9‘)
22         {
23
24             int digit=*str-‘0‘;
25             if(result<=INT_MAX/10)
26             {
27                 result*=10;
28             }
29             else
30             {
31                 if(flag==1) return INT_MAX;
32                 else return INT_MIN;
33             }
34
35             if(result<=INT_MAX-digit)
36             {
37                 result+=digit;
38             }
39             else
40             {
41                 if(flag==1) return INT_MAX;
42                 else return INT_MIN;
43             }
44
45             str++;
46         }
47
48 return flag*result;
49     }
50 };