leetcode 8 -- String to Integer (atoi)

String to Integer (atoi)

题目：
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.
Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

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题意：
字符串转化为整型，很常见的一个题，就是考虑的情况比较多而已，题目给了那么长一段话也是告诉你要考虑的情况会很多。

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思路：
考虑的情况如下：
1.” 123456” 输出：123456
2.”u123456” 输出：123456
3.”123 123” 输出：123
4.”—123456” 输出：123456
5.”+-2” 输出：-2 , 测试leetcode输出是0
6.”-+2” 输出：-2 , 测试leetcode输出是0，实际int i = +-2，输出i的确是-2。
7.”—-u123456” 输出：123456
8.”-111111111111” 输出：-2147483648
9.”111111111111” 输出：2147483647
先提取出连续的数字字符串，然后判断负号的个数，判断是否越界等等，其实也就是考虑情况多一些，我考虑的情况和leetcode有出入，并没有通过，仅供参考。

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代码：

#include <stdio.h>#include <stdlib.h>#include <assert.h>#include <string>#include <iostream>#include <cmath>using namespace std;int myAtoi(string str){    string s = "0123456789";    if(str.size() == 0)        return 0;    //截取出字符串,不应该有0    auto beg = str.find_first_of("123456789");    if(beg >= str.size())        return 0;    auto end = beg+2;    while(1)    {        if(str[end] >= 48 && str[end] <= 57)        {            ++end;        }else        {            break;        }    }    auto ret = str.substr(beg, end-beg);    int flag = 0;    //找寻负号的个数，判断正负    for(int i = 0; i <= beg; ++i){        if(str[i] == '-')        {            flag++;        }    }    //如果数字位数超过最大值位数，越界，判断正负后直接返回    if(ret.size() > 10)    {        if(flag%2 == 1){            return -2147483648;        }else{            return 2147483647;        }    }    //声明为double因为double范围大可以找到越界的情况    double d = 0;    int t = ret.size()-1;    //字符串转化为整型    for(char a:ret)    {        d += (a-48)*pow(10,t);        t--;    }    //转化正负    if(flag%2 == 1)        d = -d;    //判断是否数字位数相等且越界越界，越界返回最大或者最小值    if(d > 2147483647){        return 2147483647;    }else if(d < -2147483648){        return -2147483648;    }else{        //将double转换为int返回        return static_cast<int>(d);    }    return 0;}int main(int argc, char *argv[]){    cout << myAtoi("   123456") << endl;    cout << myAtoi("u123456") << endl;    cout << myAtoi("123 123") << endl;    cout << myAtoi("---123456") << endl;    cout << myAtoi("+-2") << endl;    cout << myAtoi("-+2") << endl;    cout << myAtoi("----u123456") << endl;    cout << myAtoi("-111111111111111") << endl;    cout << myAtoi("111111111111111") << endl;    return EXIT_SUCCESS;}

运行结果：