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，然后判断是不是正负号。最后开始计算数字，其中要计算溢出
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));
//判断是否溢出，因为溢出，数字就不是原来的数了，所以反运算之后的结果肯定不等于原来的
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);
}
//判读是否是数字
private static boolean isNumber(char ch){
if(ch >= '0' && ch <= '9')
return true;
else
return false;
}
//字符转为数字
private static int charToInteger(char ch){
return ch - '0';
}
}
思路是很简单但是在计算的过程中发现，当输入的是2147483648的时候，应该是溢出的，但是在判断溢出的那个地方竟然能过，我在Debug下看得时候，明明看到的是数字变成负数，但是不知道为什么反运算还是正确的。所以我只能加sum < 0来判断溢出，但是这样好像并不好，不知道有没有人解答一下