[LeetCode 8] String to Integer(atoi)

Describe:
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.

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.

#include<iostream>#include<cstdlib>using namespace std;class Solution{public:    /*    leetcode上以这样的函数签名也是有它很大的意义的，尽管使用const String& str,以引用作为参数避免拷贝构造函数    的调用，但这样的话会导致参数只能传递string对象，而不能传递C风格的字符串，如MyAtoi("123");这时就报错了；    */    //int StringToInteger(const string& str)    int MyAtoi(string str)    {        int size = str.size(); // 获取字符的个数        int ret = 0;         int sign = 2; // 这个初值不设置为2是有别用的,只要是大于0的即可，表示是正数        bool IsBegin = false;        for(int i=0; i<size; i++)        {            if( str[i] < '0' || str[i] > '9')            {                if(ret != 0) // 避免"12 41 21414"中间出现非数字字符                    return ret*(sign>0?1:-1);                if(str[i] == ' ' )                {                    if(IsBegin)                        return ret*(sign>0?1:-1);                    else                         continue;                }                else if(str[i] == '-' || str[i] == '+')                {                    if(sign != 2) // 避免出现"-+343"或 "+-3243" 或"++3432" ，"--132"这种情况                        return 0;                    sign = (str[i] == '-' ? -1 : 1);                    IsBegin = true;                }                else                     return 0;            }            else            {                   IsBegin = true;                // 这个条件判断很好很完善，对比以下注释的的判断条件好好领悟，感受其中的奥妙                if(ret > INT_MAX / 10 || (ret == INT_MAX /10 && (str[i]-'0')>INT_MAX%10))                     return sign > 0 ? INT_MAX : INT_MIN;                ret = ret * 10 + str[i] - '0';                 // if(sign > 0 && (INT_MAX - ret) < 0) // 上溢                // if(sign > 0 && (unsigned int)ret > INT_MAX)                // {                    // return INT_MAX;                // }                // // 下溢 类似上面结论，减就是加， 这个判断很重要，好好领会                // else if(sign < 0 && (unsigned int)(ret * sign * -1) > INT_MAX)                 // {                    // return INT_MIN;                //}            }        }        return ret*(sign>0?1:-1);    }};int main(){    Solution s;    cout << "===========================\n";    cout << "以下为正常数据测试:" << endl;    cout << s.MyAtoi(" +00123") << endl;    cout << s.MyAtoi("9") << endl;    cout << s.MyAtoi("   91") << endl;    cout << s.MyAtoi("2147483646") << endl;    cout << s.MyAtoi("    +001340") << endl;    cout << s.MyAtoi("-2147483646") << endl;    cout << s.MyAtoi("  -283646") << endl;    cout << s.MyAtoi("    -00134") << endl;    cout << s.MyAtoi("    -0134") << endl;    cout << s.MyAtoi("    -001034") << endl;    cout << "===========================\n\n";    cout << "===========================\n";    cout << "以下为溢出测试:" << endl;     cout << s.MyAtoi("    21474836450") << endl;    cout << s.MyAtoi("  -21474836492") << endl;    cout << s.MyAtoi("    10522545459") << endl; // 用以上注释的边界判断，这句不能通过测试    cout << "===========================\n\n";    cout << "===========================\n";    cout << "以下为异常字符串测试，结果返回前面那截数或0:" << endl;    cout << s.MyAtoi("32dfad") << endl;    cout << s.MyAtoi("-14gh") << endl;    cout << s.MyAtoi("++23") << endl;     cout << s.MyAtoi("--2343") << endl;    cout << s.MyAtoi(" +-123") << endl;    cout << s.MyAtoi("-[phone]") << endl;    cout << s.MyAtoi("9ad") << endl;    cout << s.MyAtoi("   +0 123") << endl;    cout << s.MyAtoi("   - 123") << endl;    cout << "===========================\n\n";}

小结：
1）刷这方面的细节题经验不足，这道题从基本功能实现到AC经历了差不多1个多小时，尚待提高，但最起码起步了就是进步了，继续加油；
2）题目中主要考察细心，考虑各种可能输入的情况，自己做题的过程是先搭建好最起码的功能实现代码，然后进行测试用例的设计，在测试当前代码的时候，写测试用例就会想到其他可能的输入情况，一点一点加上这些细枝末节，但一开始不要考虑的太多，先把基本功能实现好，然后在测试的时候就会想到越来越多的可能情况，再添加代码即可；