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.
提示： 使用string的特性来判断是否越界。 strcmp
1.去掉字符串之前多余的空格
2. 判断正负号
3. 记录数字字符串。
4. 判断是否越界。
利用好：strcmp
1 class Solution {
2 public:
3 int atoi(const char *str) {
4 const char* Max = "2147483647";
5 const char* Min = "2147483648";
6 char b[100];
7 int number=0;
8 int below=0,i=0,j=0;
9 while(str[i]==‘ ‘)i++;
10 if(i==strlen(str)) return 0;
11
12 if(str[i]==‘+‘||str[i]==‘-‘)
13 {
14 if(str[i]==‘-‘) below=1;
15 i++;
16 }
17
18 for(;i<strlen(str);i++)
19 {
20 if(str[i]>=‘0‘ && str[i]<=‘9‘)
21 b[j++]=str[i];
22 else
23 break;
24 }
25 b[j]=‘\0‘;
26 int lb = strlen(b);
27 if(lb>10)
28 {
29 if(below) return INT_MIN;
30 else return INT_MAX;
31 }
32 else if(lb==10)
33 {
34 if(below)
35 {
36 if(strcmp(b,Min)>=0) return INT_MIN;
37 else
38 {
39 for(j=0;j<lb;j++)
40 number= number*10 + b[j]-‘0‘;
41 }
42 }
43 else
44 {
45 if(strcmp(b,Max)>=0) return INT_MAX;
46 else
47 {
48 for(j=0;j<lb;j++)
49 number= number*10 + b[j]-‘0‘;
50 }
51 }
52 }
53 else
54 {
55 for(j=0;j<lb;j++)
56 number= number*10 + b[j]-‘0‘;
57 }
58 return below>0?-number:number;
59 }
60
61 };
分支结构有点多，需要仔细分析分析。
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