[Leetcode] 008. String to Integer (easy) (C++/java/python)

Index: [Leetcode] leetcode key index (C++/JAVA/PYTHON/SQL)
Github:https://github.com/illuz/leetcode

008.string_to_integer (Easy) links：

Title: https://oj.leetcode.com/problems/string-to-integer-atoi/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

Converts a string to type int.

Analysis：

Note If you go out of range, the nearest int is returned.
If: 2147483648 is greater than Int_max (2147483647), return 2147483647.

Before can use sscanf lazy, recently updated case is stuck.
Some points to note:

1. Skip the preceding spaces, \t,\n
2. Scope definition

Regular expressions using Python can be easily handled.

Code：

C++:

Class Solution {public:    int atoi (string str) {int ret = 0;bool overflow = False;int sign = 1;//default is ' + ' int i = 0;int len = Str.length (), while (I < Len && (str[i] = = ' | | str[i] = = ' \ n ' | | str[i] = = ' \ t ')) ++i;if (i = = Len return 0;//get signif (str[i] = = '-') {++i;sign =-1;} else if (str[i] = = ' + ') ++i;while (i < len) {if (!isdigit (str[i ]) break;if (sign = = 1 && ret > (Int_max-(str[i]-' 0 '))/10) | | (sign = =-1 &&-ret < (int_min + (str[i]-' 0 '))/10)) {overflow = True;break;} RET = RET * + (Str[i]-' 0 '); ++i;} if (overflow) ret = (sign = = 1)?  Int_max:int_min;elseret *= sign;return ret;    };

Java:

public class Solution {public int atoi (String str) {int ret = 0;        Boolean overflow = false;        int sign = 1;//The default is ' + ' int i = 0;        int len = Str.length ();        while (I < Len && (Str.charat (i) = = ' | | Str.charat (i) = = ' \ n ' | | str.charat (i) = = ' \ t ')) ++i;        if (i = = len) return 0;            Get sign if (Str.charat (i) = = '-') {++i;        sign =-1;        } else if (Str.charat (i) = = ' + ') ++i;            while (I < len) {if (Str.charat (i) < ' 0 ' | | str.charat (i) > ' 9 ') break;                    if (sign = = 1 && ret > (Integer.max_value-(Str.charat (i)-' 0 ')/10) | | (sign = =-1 &&-ret < (Integer.min_value + (Str.charat (i)-' 0 '))/10))                {overflow = true;            Break            } RET = ret * + (Str.charat (i)-' 0 ');        ++i;         } if (overflow)   RET = (sign = = 1)?        Integer.MAX_VALUE:Integer.MIN_VALUE;        else ret *= sign;    return ret; }}

Python:

Class solution:    # @return An integer    def atoi (self, str):        str = Str.strip ()        if not str:            return 0        Max_int = 2147483647        min_int = -2147483648        ret = 0        overflow = False        pos = 0 Sign        = 1        if Str[pos]  = = '-':            pos + = 1 Sign            = 1        elif Str[pos] = = ' + ':            pos + = 1 for        i in range (POS, len (str)):            If Not Str[i].isdigit ():                break            ret = ret * + int (str[i])            If isn't min_int <= sign * ret <= MAX_INT:
   overflow = True                break        if overflow:            return max_int if sign = = 1 Else min_int        else:            return sign * RET

Use Python's regular expression:

Class solution:    # @return An integer    def atoi (self, str):        str = str.strip ()        str = re.match (R ' ^[+-]?\d+ ', STR). Group ()        max_int = 2147483647        min_int = -2147483648        try:            ret = INT (str)            if ret > max_int:                return max_int            elif ret < min_int:                return min_int            else:                return ret        except:            return 0

[Leetcode] 008. String to Integer (easy) (C++/java/python)