An algorithm:
Please come to implement a atoi function, it can convert a string to an integer.
First, the function will begin with a space character discard useless if necessary, until the find to the first non-space character so far.
When we find the first non-space character is a positive or negative number, the combination of the symbols as much as possible with consecutive numbers up later, as the sign of integer; if the first non-space character is figures, which directly after the continuous numeric characters are combined to form an integer.
In addition to the string after a valid integer part may also exist extra characters, these characters can be ignored, they should not affect a function.
Note: if the character string in the first non-space character is not a valid integer character string is empty or contains only white space character string, then you will not need to be a function of conversion.
In any case, if the function can not effectively convert, 0 is returned.
Description:
We assume that the environment can store 32-bit signed integer size, then the numerical range [-231 231--1]. If the value exceeds this range, return INT_MAX (231 - 1) or INT_MIN (-231).
Example 1:
Input: "42" Output: 42
Example 2:
Input: "-42" Output: -42 Explanation: a first non-blank character '-', it is a negative sign. We will all digital consecutive negative number and later combined as much as possible, and finally get -42.
{Solution class
public int myAtoi (String STR) {
int m = 0;
// only the first digit encountered before or sign BREAK
for (; m <str.length (); m ++) {
char = CH STR. the charAt (m);
IF (CH == '+' || CH == '-') {
BREAK;
} the else IF (CH> = '0' && CH <= '. 9') {
BREAK;
} the else IF ( ! CH = '') {
return 0;
}
}
// will bounds determined whether m
iF (m> = str.length ()) {
return 0;
}
int RES = 0;
// K is used to indicate a positive sign -1 negative
int k = 1;
IF (str.charAt (m) == '-') {
K = -1;
m ++;
} the else IF (str.charAt (m) == '+') {
m ++;
}
the while (m <str.length ( )) {
char CH = str.charAt (m ++);
IF (CH> = '0' && CH <= '. 9') {
// the character to an integer
int A = CH - '0';
// n Analyzing whether the number of overflow
iF (K ==. 1 && (RES> Integer.MAX_VALUE / 10 || (RES Integer.MAX_VALUE == / 10 && A>. 7))) {
return Integer.MAX_VALUE;
}
iF (K == - 1 && (res * k <Integer .MIN_VALUE/ 10 || (res * k == Integer.MIN_VALUE /10 && a > 8 ))){
of Integer.MIN_VALUE return;
}
RES = RES + A * 10;
} the else {
BREAK;
}
}
// returns the result, multiplied by k (k representatives are positive or negative)
return RES * k;
}
}
2 English documentation: https://github.com/LMAX-Exchange/disruptor/wiki/Introduction This is the disruptor ah tell Java cache by using a computer program, improve project performance