LeetCode 8. String to Integer (atoi) （字符串到整数）

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 *

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题目标签：String

　　题目给了我们一个 str，让我们把它 转换为 int。

　　其中有很多违规的条件没有说明：

　　　　正负的符号只能有0个 或者 1个；

　　　　符号后面就应该是数字了，如果遇到不是数字的符号，返回目前为止合格的数字，不需要考虑后面的数字；

　　　　如果数字overflow，大于MAX的要返回MAX，小于MIN 的要返回MIN；

　　　　etc。

 

 

Java Solution:

Runtime beats 57.67% 

完成日期：01/09/2017

关键词：String

关键点：考虑到所有违规情况

class Solution
{
public int myAtoi(String str)
{
long res = 0;        // the res number to return. Note: res to return should be long and cast it to int when return it at the end.
int sign = 1;        // the sign before the number. default is 1 (positive).
int index = 0;        // index for num string to go through.

// Step 0: if parameter str is null or "", then return 0.
if(str.length() == 0 || str == null)
return 0;

// Step 1: trim the whitespace.
str = str.trim();

// Step 2: check first char is '+' or '-', move the index by 1 and also sign value.
if(str.charAt(0) == '+')
index++;
else if(str.charAt(0) == '-')
{
index++;
sign = -1;    // change the sign to -1 (negative).
}

// Step 3: go through the str string.
for(; index<str.length(); index++)
{
// if this char is not a number char, then break. No matter there are more numbers after.
if(str.charAt(index) > '9' || str.charAt(index) < '0')
break;

// add this char value into res.
res = res * 10 + (str.charAt(index) - '0');        // char - '0' is the correct int value.

// check the num exceed the max or not.
if(res > Integer.MAX_VALUE)        // res should be long because here res might be over Integer.max value.
break;
}

// Step 4: depending on the sign and max or min value, return res.
if(res * sign >= Integer.MAX_VALUE)
return Integer.MAX_VALUE;
else if(res * sign <= Integer.MIN_VALUE)
return Integer.MIN_VALUE;

// goes here meaning the res number doesn't exceed max and min integer value.
return (int)res * sign;        // here need to cast res to int.
}
}

参考资料：N/A

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