atoi

enum Status{kValid=0, KInvalid};
int g_nStatus=kValid;
int StrToInt(const char *str)
{
    g_nStatus=KInvalid;
    long long num=0;
    if((str!=NULL)&&(*str!=‘\0‘))
    {
        bool minus=false;
        if(*str==‘+‘)str++;
        else if(*str==‘-‘)
        {
            minus=true;
            str++;
        }
        if(*str!=‘\0‘)
        {
            num=StrToIntCore(str, minus);
        }
    }
    return (int)num;
}
long long StrToIntCore(const char * digit, bool minus)
{
    long long num=0;
    while(*digit!=‘\0‘)
    {
        if(*digit>=‘0‘&&*digit<=9)
        {
            int flag=minus?-1:1;
            num=10*num+flag*(*digit-‘0‘);
            if(((!minus)&&num>0x7fffffff)||(minus&&(num<(signed int)0x80000000)))
            {
                num=0;
                break;
            }
            digit++;
        }
        else
        {
            num=0;
            break;
        }
    }
    if (*digit==‘\0‘)
    {
        g_nStatus=kValid;
    }
    return num;
}

时间： 2024-10-31 23:47:59

atoi的相关文章

LeetCode：String to Integer (atoi)

1.题目名称 String to Integer (atoi) (字符串到数字的转换) 2.题目地址 https://leetcode.com/problems/string-to-integer-atoi/ 3.题目内容英文:Implement atoi to convert a string to an integer. 中文:实现atoi函数,将输入的字符串(String类型)转换为整型数据(Integer类型) 提示:实现的atoi函数需要满足以下特征忽略字符串第一个非空格字符前的所

String to Integer (atoi) leetcode

题目的意思是要将一个字符串转换成数字这个题目的重点是要处理各种各样的输入情况在题目下面有一大段英文: 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 opt

Leetcode 数 String to Integer (atoi)

本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie String to Integer (atoi) Total Accepted: 9862 Total Submissions: 67880 Implement atoi to convert a string to an integer. Hint: Carefully consider all possible input cases. If you want a challenge,

8. String to Integer (atoi)

https://leetcode.com/problems/string-to-integer-atoi/description/ 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 poss

【LeedCode】String to integer(atoi)

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 spe

【C语言】模拟实现atoi函数

atoi(表示 ascii to integer)是把字符串转换成整型数的一个函数. atoi()函数会扫描参数 nptr字符串,跳过前面的空白字符(例如空格,tab缩进等,可以通过isspace( )函数来检测),直到遇上数字或正负符号才开始做转换,而再遇到非数字或字符串结束时('\0')才结束转换,并将结果返回.如果 nptr不能转换成 int 或者 nptr为空字符串,那么将返回0 我们在模拟实现atoi函数时,要注意以下几点: 1.字符串之前的空白问题 2.正负号 3.字符串为空时 4.

LeetCode8 String to Integer (atoi)

题意: Implement atoi to convert a string to an integer. (Easy) 分析: 就是注意各种特殊情况,边界情况的判断,见代码注释. 1 class Solution { 2 public: 3 int myAtoi(string str) { 4 int start = 0; 5 long long result = 0; 6 int flag = 0; 7 //开头多余空格的处理 8 while (str[start] == ' ') { 9

[leetcode]经典算法题- String to Integer (atoi)

题目描述: 把字符串转化为整数值原文描述: 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

c++ 实现atoi()函数

1. 问题描写叙述实现c++函数库中atoi()函数,要考虑到各种特殊情况: 空字符串. +和-号. 字符串前中后n个空格. 溢出. 非数字字符. 2. 解决方式转换过程并不复杂.复杂的是要考虑到众多特殊情况. int myAtoi(string str) { if(str.length() == 0) return 0; //空串 bool isNeg = false; long re = 0; int i=0,cnt; for(;i<str.length(); i++) if(str[i

String to Integer (atoi)

1. Question 将字符串转换为整数,考虑各种输入情况: 空格处理:开头空格省略有效数字:从第一个非空格字符开始的是+.-或数字,直到下一个非数字字符结束. 加号处理:开头加号省略空串处理溢出处理无效数字处理 Implement atoi to convert a string to an integer. Hint: Carefully consider all possible input cases. If you want a challenge, please do no