gpt4 book ai didi

java - 快速排序方法 arrayOutofBounds 异常

转载 作者:行者123 更新时间:2023-12-02 08:04:23 25 4
gpt4 key购买 nike

我正在尝试使用快速排序方法来对部分或部分数据进行排序。我认为我的分区方法是正确的,我对左段和右段进行快速排序的递归调用给了我一个数组越界异常,但我似乎无法弄清楚。这是我编写的代码,并且在

中收到错误
 private void quickSorter(String[] data, int firstIndex, int numberToSort) {
if (numberToSort > 1) {
int pivot = partition(data, firstIndex, numberToSort);
int right = firstIndex + numberToSort -1;
if (firstIndex < pivot -1)
quickSorter(data, firstIndex, pivot -1);
if (pivot < right && right <data.length)
quickSorter(data, pivot , right);
}
}
private void swap(String[] data ,int tooBigNdx, int tooSmallNdx) {
String temp = data[tooBigNdx];
data[tooBigNdx] = data[tooSmallNdx];
data[tooSmallNdx] = temp;

}

@Override
public int partition(String[] data, int firstIndex, int numberToPartition) {
int ndx = firstIndex;
String pivot = data[ndx];
int tooBigNdx = ndx;
int tooSmallNdx = firstIndex + numberToPartition -1;
while (tooBigNdx < tooSmallNdx) {
while (tooBigNdx < tooSmallNdx && (data[tooBigNdx].compareTo(pivot) < 0 || data[tooBigNdx].equals(pivot)))
tooBigNdx++;

while (tooSmallNdx > ndx && (data[tooSmallNdx].compareTo(pivot) > 0 || data[tooSmallNdx].equals(pivot)))
tooSmallNdx--;
if (tooBigNdx < tooSmallNdx ) {
swap(data, tooBigNdx,tooSmallNdx);
tooBigNdx++;
tooSmallNdx--;
}
}
if (pivot.compareTo(data[tooSmallNdx]) > 0 || pivot.equals(data[tooSmallNdx]) ) {
swap(data, ndx,tooSmallNdx);
return tooSmallNdx;
} else return ndx;


}

编辑:

用于测试的数据集每次都是随机创建的,并且是5个字母的单词。我总是收到 arrayoutofBounds 异常,但超出范围的数量会根据数据集而变化。

java.lang.ArrayIndexOutOfBoundsException: -1
at sortcomparison.BasicSorter.partition(BasicSorter.java:62)
at sortcomparison.BasicSorter.quickSorter(BasicSorter.java:37)
at sortcomparison.BasicSorter.quickSorter(BasicSorter.java:40)
at sortcomparison.BasicSorter.quickSorter(BasicSorter.java:40)
at sortcomparison.BasicSorter.quickSorter(BasicSorter.java:40)
at sortcomparison.BasicSorter.quickSort(BasicSorter.java:31)
at sbccunittest.BasicSorterTester.standardSortTest(BasicSorterTester.java:166)
at sbccunittest.BasicSorterTester.testQuickSort(BasicSorterTester.java:56)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
at java.lang.reflect.Method.invoke(Unknown Source)
at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:44)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:15)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:41)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:20)
at org.junit.internal.runners.statements.RunBefores.evaluate(RunBefores.java:28)
at org.junit.internal.runners.statements.RunAfters.evaluate(RunAfters.java:31)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:263)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:69)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:48)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:231)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:60)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:229)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:50)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:222)
at org.junit.runners.ParentRunner.run(ParentRunner.java:292)
at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:50)
at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:467)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:683)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:390)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:197)

最佳答案

不应该

 int right = firstIndex + numberToSort -1;

 int right = firstIndex + numberToSort - 1 - pivot;

记住,第二个参数不是索引号,而是计数。

您的另一个选择是将其更改为索引号(我认为这会更直观。)

关于java - 快速排序方法 arrayOutofBounds 异常,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8393804/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com