Implement atoi which converts a string to an integer.
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.
Note:
Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.
Example 1:
Input: "42"
Output: 42
Example 2:
Input: " -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
Then take as many numerical digits as possible, which gets 42.
Example 3:
Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:
Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical
digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5:
Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
Thefore INT_MIN (−231) is returned.
思路：首先是去掉空格，然後判斷是不是正負號。最後開始計算數字，其中要計算溢出
class Solution {
public int myAtoi(String str) {
String string = str.trim();
int flag = 0;
int index = 0;
if(string.length()==0)
return 0;
if(string.charAt(index) == '-')
{
flag = 1;
index++;
}
else if(string.charAt(index) == '+')
{
index++;
}
int sum = 0;
int result = 0;
while(index < string.length() && isNumber(string.charAt(index)) )
{
sum = sum * 10 + charToInteger(string.charAt(index));
//判斷是否溢出，因爲溢出，數字就不是原來的數了，所以反運算之後的結果肯定不等於原來的
if(((sum - charToInteger(string.charAt(index)))/10) != result || sum < 0)
{
if(flag == 1)
return Integer.MIN_VALUE;
else
return Integer.MAX_VALUE;
}
result = result*10 + charToInteger(string.charAt(index));
index++;
}
return flag == 0 ? result:result*(-1);
}
//判讀是否是數字
private static boolean isNumber(char ch){
if(ch >= '0' && ch <= '9')
return true;
else
return false;
}
//字符轉爲數字
private static int charToInteger(char ch){
return ch - '0';
}
}
思路是很簡單但是在計算的過程中發現，當輸入的是2147483648的時候，應該是溢出的，但是在判斷溢出的那個地方竟然能過，我在Debug下看得時候，明明看到的是數字變成負數，但是不知道爲什麼反運算還是正確的。所以我只能加sum < 0來判斷溢出，但是這樣好像並不好，不知道有沒有人解答一下