Question

我无法完成一项任务。我完成了代码，必须提交作业，该作业将使用一堆测试对其进行自动评分。除了两个我都通过了。我不知道测试代码。我假设它来自此方法：

/**
 * _Part 2: Implement this method._
 *
 * @return the number of items in the list.
 */
public int size() {
    int size = 0;       //int holder for size of elements in array
    for (int i = 0; i < array.length; i++){     //loop to run through up to length of array
        if(array[i] != null){           //if data at array index has a value, increase int size
            size++;
        }
        else{           //if data at array index has no value, there are not anymore elements to check for
            break;
        }
    }
    return size;        //returning the number of elements in the array
}

这些是我得到的错误：

Starting: insertTwoItems
Failed: insertTwoItems
 Hint: Size should be 2 after two inserts! expected:<2> but was:<1>
Finished: insertTwoItems

Starting: insertThreeItems
Failed: insertThreeItems
 Hint: Size should be 2 after two inserts! expected:<2> but was:<1>
Finished: insertThreeItems

构造函数很简单：array = new String [10];

这是我将新字符串插入数组的方式

@SuppressWarnings("unchecked")
/**
 * _Part 1: Implement this method._
 *
 * Inserts a new item in the OrderedArrayList. This method should ensure
 * that the list can hold the new item, and grow the backing array if
 * necessary. If the backing array must grow to accommodate the new item, it
 * should grow by a factor of 2. The new item should be placed in sorted
 * order using insertion sort. Note that the new item should be placed
 * *after* any other equivalent items that are already in the list.
 *
 * @return the index at which the item was placed.
 */
public int insert(String item) {
    if(array.length <= size()){         //checking if string array is full
        String[] tempArray = array;     //creating a temp string array, to make this.array bigger
        array = new String[array.length*2];     //doubling the length of array

        for(int i = 0; i < tempArray.length; i++){      //putting tempArray values back into original array
            array[i] = tempArray[i];
        }
    }

    for(int i = 0; i < array.length; i++){      //loop to find location to add item into
        if(array[i] == null){       //if array index location is null, then put item in that spot.
            array[i] = item;
            return i;               //returning index location item was put at
        }
        if(item.compareTo(array[i]) < 0){       //will only insert until array index is alphabetically smaller than item
            for(int j = size()-1; j > i; j--){      //moving all items from array size down to index j
                array[j+1] = array[j];          //copying left index into right index

            }
            array[i] = item;        //once all indexes from where item should go is moved to the right, index is put in the array
            return i;               //returning index of array where item went
        }
    }

    return 1000000;     //not necessary, but cannot compile without a return outside of loops. Should not reach this
}

我不确定抑制警告的含义是什么。有人看到我的代码中有任何错误可能会影响上述测试失败吗？

Answer 1

问题出在这部分。

for(int j = size()-1; j > i; j--){      //moving all items from array size down to index j   
    array[j+1] = array[j];             //copying left index into right index
}

j的分配不正确。例如，由于您有一个数组['a','z']，并且您将要插入'c'。 j的值为1，而您的i的值为1，因此不会进入循环。您应该将其更改为此。

for (int j = size(); j >= i; j--) { // moving all items from array size down to index j
    array[j] = array[j - 1]; // copying left index into right index
}

现在j的值将为2，而i的值将为1，因此它将进入循环。然后，您将左索引复制到右索引。

与此有关。

return 1000000;

这实际上是必需的，但您的退货必须为-1。返回-1意味着未插入项目，因为-1在数组中不是有效索引。

初学者字符串数组操作

1 个答案: