Leetcode Problem.8—String to Integer (atoi)-优快云博客

本文链接：https://blog.youkuaiyun.com/xingyanxiao/article/details/46318783

本文介绍了一种使用 C++ 来实现字符串到整数转换的方法。该方法首先去除字符串前导空格，检查符号位，并遍历字符串将字符逐个转换为数字，同时考虑了整数溢出的情况。

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.

My C++ solution！

  int myAtoi(string str) 
    {
     bool flag=false;
	char label;
	int len=str.length();
	//去掉空格
	while(str[0]==32&&len)
	{
		str.erase(0,1);
		len--;
	}
		//首字母
	if(len)
	{
		if((str[0]>'9'||str[0]<'0')&&str[0]!='-'&&str[0]!='+')
			return 0;
		else if(str[0]=='-'||str[0]=='+')
		{
			if(len==1)
				return 0;
			else if(str[1]>'9'||str[1]<'0')
				return 0;
		}
	}
	if(str[0]=='-'||str[0]=='+')
	{
		label=str[0];
		flag=true;
		str=str.substr(1,len-1);
	}
	len=str.length();
	long long integer=0;


	int i=0;
	while(len&&str[i]<='9'&&str[i]>='0')
	{
		integer=integer*10+str[i]-'0';
		if(label=='-'&&integer>=0x80000000)
		{
			integer=0x80000000;
			break;
		}
		else if(integer>0x7fffffff)
		{
			integer=0x7fffffff;
			break;
		}
		len--;	
		i++;
	}
	if(flag==true)
		integer=(label=='-')?(-integer):integer;
	return integer;
    }