LeetCode #String to Integer (atoi)#

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LeetCode #String to Integer (atoi)#

自己写一个atoi. 其实之前都没怎么调用过这个库函数. 具体性能也不怎么知道.

自己写的时候感觉有个版本比题目要求还好,但是不能被接受, 于是就改改改

话说AC的感觉真爽...

我尽量把代码写的紧凑...

"""
Programmer  :   EOF
Date        :   2015.04.01
File        :   string_to_integer.py

"""

class Solution():

    INT_MAX = 2147483647
    INT_MIN = -INT_MAX - 1

    def is_digital(self, c) :
        if len(c) == 0 :
            return False

        if '9' >= c and c >= '0':
            return True

        return False
            
    def atoi(self, string) :
        length = len(string)

        if length == 0 :
            return 0

        num = 0
        found_tag = False
        positive = True
        blank_cleared = False

        for i in range(0, len(string)):

            if blank_cleared is False  and string[i] == ' ':
                "Do nothing"
                
            elif blank_cleared is True and string[i] == ' ':
                break

            elif string[i] == '+' or string[i] == '-':
                blank_cleared = True
                if found_tag is False :
                    found_tag = True

                    if string[i] == '-' :
                        positive = False
                else:
                    return 0

            elif self.is_digital(string[i]) is True:
                blank_cleared = True
                num *= 10
                num += int(string[i]) - int('0')
            else:
                break

        if positive is False :
            num *= -1

        if num > INT_MAX :
            return INT_MAX
        elif num < INT_MIN :
            return INT_MIN
        else :
            return num


s = Solution()

print s.atoi("+   123hello45")
print s.atoi("-12345")
print s.atoi("12345")
print s.atoi("    +004500")
print s.atoi("    +0 04500")

默认的数值为正数. 我用了个变量positive来标记符号. 多次遇到正负符号说明是错误的字符串. return 0 所以我用了个found_tag来标记是否我们已经找了正负符号说明, 重新遇到我们认为字符串错误.

最后做好数据范围处理, 不要超过整数范围.

好, 我们来看看皓神的作品:

// Source : https://oj.leetcode.com/problems/string-to-integer-atoi/
// Author : Hao Chen
// Date   : 2014-06-18

/********************************************************************************** 
* 
* 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. 
* 
* 
* 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.
*               
**********************************************************************************/

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

#define INT_MIN     (-2147483647 - 1)
#define INT_MAX      2147483647

int atoi(const char *str) {
    if (str==NULL || *str=='\0'){
        return 0;
    }
    
    int ret=0;
    
    for(;isspace(*str); str++);
    
    bool neg=false;
    if (*str=='-' || *str=='+') {
        neg = (*str=='-') ;
        str++;
    }
    
    for(; isdigit(*str); str++) {
        int digit = (*str-'0');
        if(neg){
            if( -ret < (INT_MIN + digit)/10 ) {
                return INT_MIN;
            }
        }else{
            if( ret > (INT_MAX - digit) /10 ) {
                return INT_MAX;
            }
        }

        ret = 10*ret + digit ;
    }
    
    return neg?-ret:ret;
}


int main()
{
    printf("\"%s\" = %d\n", "123", atoi("123"));
    printf("\"%s\" = %d\n", "   123", atoi("    123"));
    printf("\"%s\" = %d\n", "+123", atoi("+123"));
    printf("\"%s\" = %d\n", " -123", atoi(" -123"));
    printf("\"%s\" = %d\n", "123ABC", atoi("123ABC"));
    printf("\"%s\" = %d\n", " abc123ABC", atoi(" abc123ABC"));
    printf("\"%s\" = %d\n", "2147483647", atoi("2147483647"));
    printf("\"%s\" = %d\n", "-2147483648", atoi("-2147483648"));
    printf("\"%s\" = %d\n", "2147483648", atoi("2147483648"));
    printf("\"%s\" = %d\n", "-2147483649", atoi("-2147483649"));
    return 0;
}

最后我准备翻番标准库的实现的, 但是翻看之后感觉太"库化"了. 感觉各种难看. 就算了...

LeetCode #String to Integer (atoi)#

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原文地址：http://blog.csdn.net/cinmyheart/article/details/44806279