Leetcode [8. String to Integer (atoi)]

Problem：8. String to Integer (atoi)

Question

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.

思路

考虑以下几种情况

纯数字类型"012213" -> 12213字符串有其他符号情况"222.22" -> 222"abcd222" -> 0"22aba2323"-> 22带有正负号情况"+222.22" -> 222"-22" -> -22空格情况"    1212" -> 1212"  [phone]" -> 12

操作流程
1. 首先除掉字符串前面的空格
2. 如果前面有正负号，优先处理好正负号
3. 截取前面的数字部分做转化（如果字符串最前面的子串是数字，截止到第一个非数字字符）
4. 考虑是否溢出

代码

class Solution {public:    int myAtoi(string str) {        eraseEmptyCharater(str);        int flag = 1;        int overflow = false;        if (str.size() !=0 && str[0] == '-')            flag = -1;        preprocess(str);        if(str.size() == 0)            return 0;        int result = 0;        int exp = str.size()-1;        for(int i = 0; i < str.size(); i++) {            if (unsigned(result) + unsigned((int(str[i])-48) * pow(10, exp)) > INT_MAX)                overflow = true;            result += (int(str[i])-48) * pow(10, exp);            exp--;        }        if(overflow)            return flag == 1 ? INT_MAX : INT_MIN;        return result*flag;    }    void eraseEmptyCharater(string &str) {        int i = 0;        for (; i < str.size(); i++) {            if (str[i] != ' ')                break;        }        str = str.substr(i, str.size()-i);    }    void preprocess(string &str) {        if(str[0] == '+' || str[0] == '-')            str = str.substr(1, str.size()-1);        string vocabulary = "0123456789";        int lastDigPos = -1;        for(int i = 0; i < str.size(); i++) {            if (vocabulary.find(str[i]) != string::npos)                lastDigPos++;            else break;        }        str = str.substr(0, lastDigPos+1);    }};