题目：
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.
题意：
讲给定字符串从第一个非空字符起算出字符串中的整数
思路：
磨人的细节题，一开始随便写WA好几次，不断根据样例修改代码，改得一团糟，然后推倒重来认真审题才发现我太年轻。正确的代码不会太过繁琐杂乱。成功AC，Runtime: 20 ms, faster than 95.89% of C++ online submissions for String to Integer (atoi)~~~
1、从字符串第一个非空字符开始；
2、若第一个非空字符不为‘+’、‘-’或‘0‘~’9’之间的字符返回0；
3、‘+’和‘-’只能有一个，若同时存在返回0；
4、碰到非数字字符，结束字符串遍历；
5、字符串中数字可能有很多很多位，甚至超过long long的范围，故在遍历数字字符时边遍历边判断当前形成的整数有没有超过给定的范围[−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;
}
};