LeetCode 007 008 190 To be continue

7 Reverse Integer

1. int reverse(int x) {
   
2.     long long ans=0;
   
3.     const int maxint=0x7fffffff;
   
4.     const int minint=0x80000000;
   
5.     do(x!=0){
   
6.         ans=ans*10+(x%10);//对正数负数都能处理
   
7.         x/=10;
   
8.     }while(x!=0);
   
9.     if(ans<minint || ans>maxint){
   
10.         ans=0;
    
11.     }
    
12.     return ans;
    
13. }

008

1. int myAtoi(char* str) {
   
2.     int i=0,sign=0;
   
3.     long long n=0;//非常重要，一定要定义long long
   
4.     const int maxint=0x7FFFFFFF;
   
5.     const int minint=0x80000000;
   
6.     const long long max=0x100000000;//非常重要，绝对值要大于minint的绝对值
   
7.     while(str[i]==' ' || str[i]=='\n' || str[i]=='\t')
   
8.         i++;
   
9.     if(str[i]=='+'){
   
10.         i++;
    
11.     }
    
12.     else{
    
13.         if(str[i]=='-'){
    
14.             sign=1;
    
15.             i++;
    
16.         }
    
17.     }
    
18.     
    
19.     for(;str[i]!='\0';i++){
    
20.         if(str[i]<='9' && str[i]>='0'){
    
21.             n=n*10+(str[i]-'0');
    
22.             if(n>max)n=max;//非常重要
    
23.         }else{
    
24.             break;
    
25.         }
    
26.     }
    
27.     if(sign)n=-n;
    
28.     if(n>maxint)n=maxint;
    
29.     if(n<minint)n=minint;
    
30.     return n;
    
31. }
32. /*
33. "9223372036854775809"
34. "2147483648" 
35. "-2147483648"
36. " -11919730356x"
    
37. */