1

我应该通过线性探测来实现我自己的哈希表，但是我在使用 remove 函数时遇到了问题。它无法正常工作，但我无法查明问题：

public void remove(Student s){

    if(s==null) throw new NullPointerException("Student is null");  /*check if student is null*/

    if (!contains(s)){                                               /*stop if students is not in table*/
        throw new NullPointerException();
    }

    int i = hashFunction(s);
    for(int j = 0; j<array.length; j++){
        if(s.equals(array[j])){                                              /*found student*/
            array[j]=null;                                                  /*delete student*/
            break;
        }
    }

    for(int k= i; k<array.length; k++){             /* find out if next element exists, if yes, move it to the gap of deleted student */
        if(!(array[k]==null)){
        int tempnext= hashFunction(array[k]);

            if(i <= tempnext){
                array[i]=array[tempnext];
            }
        }
    }   
}

我的调试器将此显示为输出，因此很明显我的方法存在表格中的空白问题：

Start:

Array-Indices: [ 0] [ 1] [ 2] [ 3] [ 4] [ 5] [ 6] [ 7]
Hash-Values:   [ 6] [ 1] [__] [ 6] [ 0] [ 5] [ 3] [ 0]
Number:        [ 9] [ 6] [__] [ 2] [ 3] [ 4] [ 5] [ 8]

Removing s6

Array-Indices: [ 0] [ 1] [ 2] [ 3] [ 4] [ 5] [ 6] [ 7]
Hash-Values:   [ 6] [ 6] [__] [ 6] [ 0] [ 5] [ 3] [ 0]
Number:        [ 9] [ 2] [__] [ 2] [ 3] [ 4] [ 5] [ 8]

Removing s5

Array-Indices: [ 0] [ 1] [ 2] [ 3] [ 4] [ 5] [ 6] [ 7]
Hash-Values:   [ 6] [ 6] [__] [ 5] [ 0] [ 5] [__] [ 0]
Number:        [ 9] [ 2] [__] [ 4] [ 3] [ 4] [__] [ 8]

提前感谢您的帮助！

4

0 回答 0