Leetcode-8 String to Integer (atoi)



Problem Description:

Implement atoi To Convert a string to an integer.

Hint: Carefullyconsider all possible input cases. If you want a challenge, please don't seebelow and ask yourself what is the possible input cases.

Notes: It isintended for the problem to be specified vaguely (ie, no given input specs). You is responsible to gather all the input requirements up front.

update (2015-02-10):
The signature of the c++ Functionhad been updated. If you still see your function signature accepts a const char * argument, please click the reload buttonto reset your code definition.

Spoilers Alert...click to show requirements for atoi.

Requirements for Atoi:

The Functionfirst discards as many whitespace characters as necessary until the firstnon-whitespace character is found. Then, starting from the character, takes anoptional initial plus or minus sign followed by as many numerical digits Aspos Sible, and interprets them as a numerical value.

The Stringcan contain additional characters after those that form the integral Number,which is ignored and has no effect The behavior of this function.

If the firstsequence of non-whitespace characters in Str are not a valid integral number, Orif No such sequence exists Beca Use either str are empty or it contains onlywhitespace characters, no conversion is performed.

If no validconversion could be performed, a zero value is returned. If the correct valueis out of the range of representable values, Int_max (2147483647) or int_min ( -2147483648) is returned.

Code: Mainly consider a variety of situations can:

Refer to C + + atoi:http://blog.csdn.net/woliuyunyicai/article/details/44101881

public class Solution {public int atoi (String str) {if (str = = NULL | | str.length () = = 0) {return 0;} char[] Str_arr = Str.tochararray (); Boolean flag = true;//flag Positive negative number, positive number is truelong result = 0;int index = 0;final long Max_int = (long) (Math.pow (2, 31)-1);//In Java you can use Integer.max_value instead of the final long Min_int = (long) (-(Math.pow (2))); INT data_length = 0 ;//record length, prevent long also overflow case//Consider the case that the preceding character is a space, such as "010", eliminate the whitespace can use String.Trim () while (Index < str.length () && str_arr[ Index] = = ") {index++;} First determine whether the first character is a positive or negative "+,-", Mark if (str_arr[index] = = ' + ') {flag = True;++index;} else if (str_arr[index] = = '-') {flag = False;++index;} Is ' 0 '-' 9 ' character while (Index < str.length () && Str_arr[index] <= ' 9 ' && Str_arr[index] >= ' 0 ' & & data_length++ <=)//data_length is used to control the number string too long {result = result * + (str_arr[index++]-' 0 ');} determine if int crosses result = flag? Result:-result;if (Result > Max_int) {result = Max_int;} else if (Result < Min_int) {result = Min_int;}  return (int) result;   } public static void Main (string[] args) {//test case//String str = "+004500";//String str = "-0012    A42 ";//String str =" 2147483648 ";//String str =" 2147483648 ";    String str = "9223372036854775809";    SYSTEM.OUT.PRINTLN (New Solution (). atoi (str)); }}

Leetcode-8 String to Integer (atoi)