排序有效.它在数组上递归执行快速排序,直到它需要排序30个元素或更少的数组,在这种情况下,它执行插入排序.
一切都很好10 ^ 3和10 ^ 4但是一旦我达到10 ^ 5的值,它只会对随机,几个独特和随机列表进行排序,但在排序几乎排序和排序的列表时会产生堆栈溢出错误.
我不相信问题在于生成列表的方式,因为堆栈溢出发生在排序算法中(编译器引用的行是findPivot()方法中的第一行).此外,它通常会在崩溃之前对1到6个列表进行排序.因此,必须以某种方式使算法本身与几乎排序和排序的列表进行交互.此外,反向列表的生成涉及调用用于创建排序列表的代码(然后将其反转).
此外,我发现该问题不太可能只是由于某种原因,算法必须通过在几乎排序和排序的列表中递归而不是其他列表类型来调用数组的部分分区,如它可以对10 ^ 7值的随机列表进行排序,这将需要比具有10 ^ 5值的近似排序列表更多的分区.
我意识到它必须与这些列表类型如何与我的快速排序的递归相互作用有关,但如果有人可以阐明它会很棒.我把代码放到了完整的快速排序算法和下面的随机列表生成器中.
快速排序算法
/** * Performs a quick sort given the indexes of the bounds of an integer array * @param arr The array to be sorted * @param highE The index of the upper element * @param lowE The index of the lower element */ public static void quickSort(int[] arr,int highE,int lowE) { //Only perform an action if arr.length > 30,otherwise insertion sort [recursive base case]) if (lowE + 29 < highE) { //Get the element and then value of the pivot int pivotE = findPivot(arr,highE,lowE); int pivotVal = arr[pivotE],storeE = lowE; //Swap the pivot and the last value. swapElements(arr,pivotE,highE); //For each element in the selection that is not the pivot,check to see if it is lower than the pivot and if so,move it to the leftmost untouched element. for (int i = lowE; i < highE; i++) { if (arr[i] < pivotVal) { swapElements(arr,storeE,i); //Increment storeE so that the element that is being switched moves to the right location storeE++; } } //Finally swap the pivot into its proper position and recrusively call quickSort on the lesser and greater portions of the array swapElements(arr,highE); //Lesser quickSort(arr,storeE - 1,lowE); //Greater quickSort(arr,storeE + 1); } else { insertSort(arr,lowE); } } /** * Finds the pivot element * @param arr The array to be sorted * @param highE The index of the top element * @param lowE The index of the bottom element * @return The index of the pivot. */ public static int findPivot(int[] arr,int lowE) { //Finds the middle element int mid = (int) Math.floor(lowE + (highE - lowE) / 2); //Returns the value of the median of the first,middle and last elements in the array. if ((arr[lowE] >= arr[mid]) && (arr[lowE] >= arr[highE])) { if (arr[mid] > arr[highE]) {return mid;} else {return highE;} } else if ((arr[mid] >= arr[lowE]) && (arr[mid] >= arr[highE])) { if (arr[lowE] > arr[highE]) {return lowE;} else {return highE;} } else { if (arr[lowE] > arr[mid]) {return lowE;} } return mid; } /** *Performs an insertion sort on part of an array * @param arr The array to be sorted. * @param highE The index of the top element. * @param lowE The index of the low element. */ public static void insertSort(int[] arr,int lowE) { //Sorts elements lowE to i in array,with i being gradually incremented. for (int i = lowE + 1; i <= highE; i++) { for (int j = i; j > lowE; j--) { if (arr[j] < arr[j - 1]) { swapElements(arr,j,j-1); } else {break;} } } }
随机列表发电机
/** * Creates a random list * @param arr The array to place the list inside of */ public static void randomList(int[] arr) { //Places a random number at each element of the array for (int i = 0; i < arr.length; i++) { arr[i] = (int) Math.floor(Math.random() * RAND_MAX); } } /** * Creates a nearly sorted list of random numbers * @param arr the array to place the list inside of */ public static void nearSortList(int[] arr) { //Creates a sorted list in arr sortList(arr); int swaps = (int) (Math.ceil(Math.pow((Math.log(arr.length)),2.0))); //The two values to be switched each time int a,b; //Performs a number of swaps equal to swaps [log(N) ^ 2] rounded up,with numbers switched no more than ln(N) places away for (int i = 0; i < swaps; i++) { a = (int) Math.floor(Math.random() * arr.length); b = (int) (a + Math.random() * 2 * Math.log(arr.length) - Math.log(arr.length)); //Accounts for cases in which b is either greater or smaller than the array bounds if (b < 0) { b = -b; } else if (b >= arr.length) { b = -1 * (arr.length - b); } swapElements(arr,a,b); } } /** * Creates a random list with many unique values in * @param arr the array to place the list inside of */ public static void fewUniqueList(int[] arr) { int[] smallArr = new int[(int) Math.floor(Math.pow(arr.length,9.0 / 10.0))]; //Creates a smaller array of random values randomList(smallArr); //Fills the larger list up with values from the smaller list,ensuring aproximately N / N ^ (9/10) instances of each number in the array and ensuring,at most,there are N ^ (9/10) (rounded down) unique values in the large array for (int i = 0; i < arr.length; i++) { arr[i] = smallArr[(int) Math.floor(Math.random() * smallArr.length)]; } } /** * Creates a reversed list of random numbers * @param arr the array to place the list inside of */ public static void reversedList(int[] arr) { //Creates a sorted list in arr sortList(arr); //Switches each ith elements with its mirror on the other end of the array until the value of i reaches the middle of the array for (int i = 0; i < (int) (arr.length / 2.0); i++) { swapElements(arr,i,arr.length - 1 - i); } } /** * Creates a sorted list of random numbers using a merge sort * @param arr the array to place the list inside of */ public static void sortList(int[] arr) { //Creates a random list in arr randomList(arr); Arrays.sort(arr); }
编辑:[已解散]
编辑2:
我已经使用以下代码替换了基本的递归调用,以便仅在EJP建议时调用两个分区中的最小分区,但仍然没有解决问题.
if (storeE - 1 - lowE < highE - storeE + 1) { //Lesser quickSort(arr,lowE); //Greater quickSort(arr,storeE + 1); } else { //Greater quickSort(arr,storeE + 1); //Lesser quickSort(arr,lowE); }
编辑3:
好吧,现在显而易见的是,递归深度远远大于我对于近乎排序和排序的列表所给予的信任.但现在我需要弄清楚为什么会出现这种情况,以及为什么随机列表的深度仅为10 ^ 7的值,但几乎排序和排序的列表的深度超过3000
解决方法
但是对于一个(几乎)排序的数组,你一次主要是分割1个元素.
因此,对于排序数组,您的堆栈大小最终会与数组的大小相同,而对于随机数组,它更可能是该大小的对数.
因此,即使随机数组比近似排序的数组大得多,因此较小的一个引发异常也就不足为奇了,但较大的一个则不然.
就修复方面而言,如EJP pointed out,您应首先执行较小的分区以限制堆栈增长.但这本身不会解决问题,因为Java doesn’t support tail-call optimization(嗯,它是可选的实现,因为我理解这个问题).
这里一个相当简单的解决方法是将函数抛入while循环,本质上是对尾调用优化进行硬编码.
为了更好地了解我的意思:
public static void quickSort(int[] arr,int lowE) { while (true) { if (lowE + 29 < highE) { ... quickSort(arr,lowE); // not doing this any more //quickSort(arr,storeE + 1); // instead,simply set the parameters to their new values // highE = highE; lowE = storeE + 1; } else { insertSort(arr,lowE); return; } } }
好吧,既然你已经有了基本的想法,那么看起来会更好(功能上与上面相同,只是更简洁):
public static void quickSort(int[] arr,int lowE) { while (lowE + 29 < highE) { ... quickSort(arr,lowE); lowE = storeE + 1; } insertSort(arr,lowE); }
这当然实际上并不首先做较小的一个,但我会留给你弄清楚(似乎你已经对如何做到这一点有了一个很好的想法).
这是如何工作的
对于一些弥补价值……
您当前的代码执行此操作:(缩进表示该函数调用内发生的事情 – 因此增加缩进意味着递归)
quickSort(arr,100,0) quickSort(arr,49,0) quickSort(arr,24,0) insertion sort quickSort(arr,26) insertion sort quickSort(arr,51) quickSort(arr,76,74) insertion sort
quickSort(arr,0) break out of the while loop insertion sort lowE = 26 break out of the while loop insertion sort lowE = 51 run another iteration of the while-loop quickSort(arr,0) break out of the while loop insertion sort lowE = 74 break out of the while loop insertion sort
增加堆栈大小
不确定你是否考虑过这个问题,或者它是否适用于你的参数,但你总是可以考虑increasing the stack size with the -Xss
command-line parameter.