leetcode 8. String to Integer (atoi) 字符串转数字

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.

题意不说了，思想也很简单，但是要注意处理各种可能出现的exception，这道题考察的主要就是对各种exception的判断和处理。

代码如下：

import java.util.ArrayList;


public class Solution 
{
     public int myAtoi(String str)
     {

         if (str == null || str.length() < 1)
                return 0;

            // trim white spaces
            str = str.trim();

            boolean flag = true;

            // check negative or positive
            int i = 0;
            if (str.charAt(0) == '-') 
            {
                flag = false;
                i++;
            } else if (str.charAt(0) == '+') 
            {
                flag=true;
                i++;
            }
            // use double to store result
            double result = 0;

            // calculate value
            while (str.length() > i && str.charAt(i) >= '0' && str.charAt(i) <= '9') 
            {
                result = result * 10 + (str.charAt(i) - '0');
                i++;
            }

            if (flag == false)
                result = -result;

            // handle max and min
            if (result > Integer.MAX_VALUE)
                return Integer.MAX_VALUE;

            if (result < Integer.MIN_VALUE)
                return Integer.MIN_VALUE;

            return (int) result;
     }


    public static void main(String[] args) 
    {
        Solution solution=new Solution();
        System.out.println(solution.myAtoi("2147483648"));
    }
}

C++做法如下：

#include <iostream>
#include <string>
#include <limits>
using namespace std;

/*
    下面的做法是做不到AC的，但是可以处理各种情况
    这道题最麻烦的就是各种乱七八糟的输入字符串
*/
class Solution 
{
public:
    int myAtoi(string str)
    {
        return atoi(str.c_str());
    }
};

下面是AC的代码：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>
#include <cstdlib>
#include <ctime>

using namespace std;


class Solution 
{
public:
    int myAtoi(string str) 
    {
        if (str.empty()) 
            return 0;

        int sign = 1, base = 0, i = 0, n = str.size();
        while (i < n && str[i] == ' ') 
            ++i;

        if (str[i] == '+' || str[i] == '-')
            sign = (str[i++] == '+') ? 1 : -1;

        while (i < n && str[i] >= '0' && str[i] <= '9')
        {
            if (base > INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - '0' > 7))
                return (sign == 1) ? INT_MAX : INT_MIN;

            base = 10 * base + (str[i++] - '0');
        }
        return base * sign;
    }
};