MergeSort算法中的Java IndexOutOfBoundsException

Question

我不断得到

线程“main”中出现异常 java.lang.ArrayIndexOutOfBoundsException:MergeSorter.merge(MergeSorter.java:44)、MergeSorter.sort(MergeSorter.java:16)、MergeSorter.sort(MergeSorter.java:14) 处的索引 2 超出长度 2 的范围，MergeSorter.sort(MergeSorter.java:14), MergeSorter.sort(MergeSorter.java:14)

不知道如何修复。

之后还想转换为通用。

public class MergeSorter {
    ////change back to generic later
    ///array is item
    public static void sort(int[] array, int begIndx, int endIndx) {
        if (array == null) {
            throw new IllegalArgumentException("Item is null.");
        }
        if(begIndx < endIndx) {
            int midIndx = (int) Math.floor((begIndx + endIndx)/2);
            sort(array, begIndx, midIndx);
            sort(array, midIndx + 1, endIndx);
            merge(array, begIndx, midIndx, endIndx);
        }
    }

    //Takes to sorted arrays and merges them together
    ///Change type of array to generic later
    public static void merge(int[] array, int begIndx, int midIndx, int endIndx) {
        int sizeOfLeft = midIndx - begIndx + 1;
        int sizeOfRight = endIndx - midIndx;

        ///change to generic later
        int[] leftArr = new int[sizeOfLeft + 1];
        int[] rightArr = new int[sizeOfRight + 1];

        //removing equal sign from loop does nothing
        for(int i = 1; i <= sizeOfLeft; i++) {
            leftArr[i] = array[begIndx + i - 1];
        }
        for( int j = 1; j <= sizeOfRight; j++) {
            rightArr[j] = array[midIndx + j];
        }
        leftArr[sizeOfLeft + 1] = Integer.MAX_VALUE;
        rightArr[sizeOfRight + 1] = Integer.MAX_VALUE;

        int i = 1;
        int j = 1;

        for(int k = begIndx; k < endIndx; k++) {
            //use comparable here
            if(leftArr[i] <= rightArr[j]) {
                array[k] = leftArr[i];
                i = i + 1;
            }else {
                ///just replaces it so don't use comparable
                array[k] = rightArr[j];
                j = j + 1;
            }
        }       
    }   
}

Answer 1

数组索引始终从零开始，因此如果您想访问数组中的第二个元素，则需要提供索引 1。

如果您想按单个值扩展数组，您可以按如下方式创建一个临时数组:

public int[] expand(int[] arrayIn) {
    int[] temp = new int[arrayIn.length + 1];
    for (int i = 0; i < arrayIn.length; i++) {
        temp[i] = arrayIn[i];
    }
    temp[arrayIn.length] = -1; // You can replace this with another "blank" value
}

因此，返回一个具有扩展索引读数的新数组(在本例中为-1。