String to Integer (atoi)

最新推荐文章于 2025-08-10 21:28:09 发布

临江别寄

最新推荐文章于 2025-08-10 21:28:09 发布

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分类专栏：算法文章标签： Java 算法

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本文介绍了一个Java实现的atoi方法，用于将输入的字符串转换为整数。该方法考虑了多种输入情况，包括空格处理、符号位判断、非法字符截断等，并进行了异常处理确保结果的有效性和正确性。

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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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

//我使用Java实现的

public int myAtoi(String str) {
// str = str.replace(" ", "");

char[] arr = str.toCharArray();
StringBuffer sb = new StringBuffer();
boolean flag = false;
for(int i = 0;i < arr.length;i++)
{
if(arr[i] == ' ' && flag == false)
{
continue;
}
else if(arr[i] >= '0' && arr[i] <= '9' || arr[i] == '-' || arr[i]=='+')
{
flag = true;
sb.append(arr[i]);
continue;
}
else
{
break;
}
}
str = sb.toString();
int ret = 0;
try
{
ret = Integer.parseInt(str);
}
catch(Exception ex)
{
if(str.equals("") || str.equals(" "))
{
ret = 0;
}
else if(str != null && (str.length() > 8))
{

try {
long longType = Long.parseLong(str);
if( (longType > 0) && (longType > Integer.MAX_VALUE))
{
ret = 2147483647;
}
else if(longType < 0 && longType < Integer.MIN_VALUE)
{
ret = -2147483648;
}
else
{
ret = (int)longType;
}
}catch(Exception e)
{
if(str.contains("-"))
{
ret = -2147483648;
}
else
{
ret = 2147483647;
}
}
}

}
finally
{

}
return ret;
}