題目：

Implement atoi which converts a string to an integer.

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.

Note:

• Only the space character ' ' is considered as whitespace character.
• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−231) is returned.

題意：

講給定字串從第一個非空字元起算出字串中的整數

思路：

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1、從字串第一個非空字元開始；

2、若第一個非空字元不為‘+’、‘-’或‘0‘~’9’之間的字元返回0；

3、‘+’和‘-’只能有一個，若同時存在返回0；

4、碰到非數字字元，結束字串遍歷；

5、字串中數字可能有很多很多位，甚至超過long long的範圍，故在遍歷數字字元時邊遍歷邊判斷當前形成的整數有沒有超過給定的範圍[−2^31,  2^31 − 1]。

Code：

class Solution {
public:
    int myAtoi(string str){
        int len=str.length();                                       
        int flag=0,i=0;
        while(str[i]==' ') i++;
        if(str[i]!='+'&&str[i]!='-'&&(str[i]<'0'||str[i]>'9')) return 0;
        if(str[i]=='-'){
            flag=1;
            i++;
        }
        else if(str[i]=='+') i++;
        long long y=0,maxr=((long long)1<<31)-1,minr=-(1<<31);
        while(str[i]>='0'&&str[i]<='9'){
            y=y*10+str[i]-'0';
            i++;
            if(y>maxr&&!flag) return maxr;
            if(y>-minr&&flag) return minr;
            if(i==len) break;
        }
        if(flag) y=-y;
        return y;
    }
};