LeetCode 007 008 190 To be continue

7 Reverse Integer

int reverse(int x) {
    long long ans=0;
    const int maxint=0x7fffffff;
    const int minint=0x80000000;
    do(x!=0){
        ans=ans*10+(x%10);//对正数负数都能处理
        x/=10;
    }while(x!=0);
    if(ans<minint || ans>maxint){
        ans=0;
    }
    return ans;
}

008

int myAtoi(char* str) {
    int i=0,sign=0;
    long long n=0;//非常重要，一定要定义long long
    const int maxint=0x7FFFFFFF;
    const int minint=0x80000000;
    const long long max=0x100000000;//非常重要，绝对值要大于minint的绝对值
    while(str[i]==' ' || str[i]=='\n' || str[i]=='\t')
        i++;
    if(str[i]=='+'){
        i++;
    }
    else{
        if(str[i]=='-'){
            sign=1;
            i++;
        }
    }

    for(;str[i]!='\0';i++){
        if(str[i]<='9' && str[i]>='0'){
            n=n*10+(str[i]-'0');
            if(n>max)n=max;//非常重要
        }else{
            break;
        }
    }
    if(sign)n=-n;
    if(n>maxint)n=maxint;
    if(n<minint)n=minint;
    return n;
}
/*
"9223372036854775809"
"2147483648"
"-2147483648"
"      -11919730356x"
*/