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java - 快速排序越界错误

转载 作者:行者123 更新时间:2023-12-01 09:58:30 25 4
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我正在尝试构建一个快速排序函数,该函数使用三个值的中位数作为基准,但我遇到了问题。我在第 56、66、35、34 和 84 行遇到越界错误,并且尝试更改此设置但无济于事。有什么想法吗?

public class MyQuickSort {

private int array[];
private int length;

public void sort(int[] inputArr) {

if (inputArr.length == 0) {
return;
}
this.array = inputArr;
length = inputArr.length;
quickSort(0, length - 1);
}

private void quickSort(int lowerIndex, int higherIndex) {

int i = lowerIndex;
int j = higherIndex;
// calculate pivot number, I am taking pivot as middle index number
int mid = array[(higherIndex-1)/2];
// Divide into two arrays

int pivot = findMedian(array, i, mid, j);
swap(pivot, 0);

while (i <= j) {
/**
* In each iteration, we will identify a number from left side which
* is greater then the pivot value, and also we will identify a number
* from right side which is less then the pivot value. Once the search
* is done, then we exchange both numbers.
*/
while (array[i] < mid) {
i++;
}
while (array[j] > mid) {
j--;
}
if (i <= j) {
swap(i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort() method recursively
if (lowerIndex < j)
quickSort(lowerIndex, j);
if (i < higherIndex)
quickSort(i, higherIndex);
}

public static int findMedian (int[] array, int a, int b, int c) {
if ((array[a] > array[b] && array[a] < array[c]) || (array[a] > array[c] && array[a] < array[b])) {
return a;
} else if ((array[b] > array[a] && array[b] < array[c]) || (array[b] > array[c] && array[b] < array[a])) {
return b;
}
return c;
}


private void swap(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}

public static void main(String a[]){

MyQuickSort sorter = new MyQuickSort();
int[] input = {24,2,45,20,56,75,2,56,99,53,12};
sorter.sort(input);
for(int i:input){
System.out.print(i);
System.out.print(" ");
}
}
}

最佳答案

问题在于变量“mid”既用作索引又用作值。函数 findMedian() 仅需要索引,而不需要数组中的值。

如果我将快速排序()更改为(请注意,这还包括 midIndex 的索引边界检查):

    private void quickSort(int lowerIndex, int higherIndex) {

int i = lowerIndex;
int j = higherIndex;
// calculate pivot number, I am taking pivot as middle index number
int midIndex = (higherIndex-1)/2;
if (midIndex < lowerIndex) {
midIndex = lowerIndex;
}
int mid = array[midIndex];
// Divide into two arrays

int pivot = findMedian(array, i, midIndex, j);
swap(pivot, i);

while (i <= j) {
/**
* In each iteration, we will identify a number from left side which
* is greater then the pivot value, and also we will identify a number
* from right side which is less then the pivot value. Once the search
* is done, then we exchange both numbers.
*/
while (array[i] < mid) {
i++;
}
while (array[j] > mid) {
j--;
}
if (i <= j) {
swap(i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort() method recursively
if (lowerIndex < j)
quickSort(lowerIndex, j);
if (i < higherIndex)
quickSort(i, higherIndex);
}

我得到:

java MyQuickSort
2 2 12 20 24 45 53 56 56 75 99

关于java - 快速排序越界错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36994900/

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