String to Integer (atoi)-LeetCode

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.

spoilers alert... click to show requirements for atoi.

Requirements for atoi: 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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

AC感想：这题目绝对是个坑，假如想挑战自己不看requirements的话。

思路：

（1）自左至右找到首个非空白符（位置记为begin），如果该非空白符为‘+’、‘-’或者 数字，则输入合法，否则非法输出0；

（2）从首个非空白符向后找非数字字符（位置记为end）。

（3）用begin和end之间的合法字符构造数字。

坑，敲黑板！！！

假如使用比较大的long来保存直接构造的数字，很不幸这个数字会大到long也溢出。那么就不能盲目的直接构造了，可以考虑

Reverse Integer（http://blog.csdn.net/kinghmy/article/details/56279266）

这道题中使用的“溢出监测”构造数字，保证构造得到的数字绝对值不会在lang下溢出，也可以最大幅度构造（或者直接拿构造出来的 数字-1 和Integer.MAX_VALUE比较也行）。

public class Solution {public int myAtoi(String str) {  long result=0;char[] number=str.toCharArray(); int loc=0,begin=0,end=number.length;while(loc<number.length){if(number[loc]==' '){loc+=1;}else break;}if(loc<number.length && (number[loc]=='-' || number[loc]=='+' ||(number[loc]>='0' && number[loc]<='9'))){begin=loc;loc+=1;while(loc<number.length){if(number[loc]>='0' && number[loc]<='9')loc+=1;else{end=loc;break;}}}else return 0;if(number[begin]=='-' || number[begin]=='+')loc=begin+1;else loc=begin;long newresult=0;while(loc<end){int tear=number[loc]-'0';newresult=result*10+tear;if(newresult/10!=result){break;}result=newresult;loc+=1;}if(number[begin]=='-')result*=-1;if(result>Integer.MAX_VALUE)return Integer.MAX_VALUE;else if(result<Integer.MIN_VALUE)return Integer.MIN_VALUE;else return (int)result;}}