在Java中，使用null字符更改char数组的大小

Question

The following code (from "Cracking the code interview", from Gaale Laakman), shows how to remove duplicate characters in a char array without using a copy of the array to avoid some extra memory use. It re-writes the final characters in the first array with an offset. Since the final array is smaller than the previous one, a null character is set at the position following the final characters, as if to say the array stops there:

    str[tail] = 0;

I was wondering if by doing this the variable "length" of the array gets changed. If not, I don't understand why this example works. Or is this just an example where we would check where is the null character to find the length of the array and don't use the length variable in question?

Here is the whole code:

    public static void removeDuplicates(char[] str) {
        if (str == null) return;
        int len = str.length;
        if (len < 2) return;
        int tail = 1;
        for (int i = 1; i < len; ++i) {
            int j;
            for (j = 0; j < tail; ++j) {
                if (str[i] == str[j]) break;
            }
            if (j == tail) {
                str[tail] = str[i];
                ++tail;
            }
        }
        str[tail] = 0;
    }

Answer 1

It sounds like a question that has been translated from C or C++. In those langauges you use a null character for the end of a string (which in turn is a char array). In Java this does not work; the array never changes its length.

If the caller knows that this null character is inserted, they can use the information of course, and ignore characters after the null. They cannot use the len variable since this only lives inside the method and does not exist when the method returns.

In Java you would usually do:

str = Arrays.copyOf(str, tail);

This would create a new array the right length and copy all the characters (which is what the code example aimed at avoiding).

By the way, I get an ArrayIndexOutOfBoundsException in the line str[tail] = 0; in the end if no duplicates were found. In this case tail is equal to the length of the array and therefore 1 position beyond the last element.

Answer 2

An array has a fixed length on creation. In the example they want to save some time by always re-using the same array for every iteration. As it is not possible to shrink the array (as the length is determined on creation), they use a work around, they put a zero at the place where the array should end. When their loop reaches the zero, it knows it is at the conceptual 'end' of the array.

Answer 3

Array are immutable so the length does not change the empty space is filled with null values

public class MainClass {

public static void main(String[] args) {
char[] org={'a','b','b','c'};
System.out.println(org.length);
System.out.println(org);
removeDuplicate(org);
System.out.println(org.length);
   System.out.println(org);

}
public static void removeDuplicate(char[]str){
if(str==null)return;
int len=str.length;
if(len<2)return;
int tail=1;
for(int i=1;i<len;++i){
    int j;
    for(j=0;j<tail;++j){
        if(str[i]==str[j])break;
    }
    if(j==tail){
    str[tail]=str[i];
    ++tail;
    }
}
   str[tail]=0;
  }
 }

**Results**
   4
  abbc
   4
  abc