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我正在编写的程序是使用堆栈实现为 QuickSort 类中的快速排序提供非递归实现。我觉得我的代码在 sort() 方法中是正确的。由于实现了 Comparable 接口,我遇到的一个问题是初始化堆栈。当我的方法具有“扩展 Comparable”时,我的 Stack 应该被参数化为什么,因为在这种情况下 E 是 Stack 的错误参数。

   package edu.csus.csc130.spring2017.assignment2;
   import java.util.Stack;
   public class QuickSort{

       // provide non-recursive version of quick sort
       // hint: use stack to stored intermediate results
       // java.util.Stack can be used as stack implementation
       public static <T extends Comparable<T>> void sort(T[] a) {
           Stack<E> stack = new Stack<E>(); //something wrong will <E> i think
           stack.push(0);
           stack.push(a.length);
           while (!stack.isEmpty()) {
               int hi = (int) stack.pop();
               int lo = (int) stack.pop();
               if (lo < hi) {
               // everything seems good up til here
               int p = partition(a, lo, hi);
               stack.push(lo);
               stack.push(p - 1);
               stack.push(p + 1);
               stack.push(hi);

               }
           }        
           throw new UnsupportedOperationException();
       }

       // Partition into a[lo..j-1], a[j], a[j+1..hi]
       private static <T extends Comparable<T>> int partition(T[] a, int lo, int   hi)    { 
           int i = lo, j = hi + 1; // left and right scan indices
           T v = a[lo]; // the pivot

           while (true) { // Scan right, scan left, check for scan complete, and exchange
                while (SortUtils.isLessThan(a[++i], v)) {//++i is evaluated to i+1 
                   if (i == hi) {
                        break;
                   }
               }
               while (SortUtils.isLessThan(v, a[--j])) {//--j is evaluated to j-1
                   if (j == lo) {
                       break;
                   }
               }
               if (i >= j) {
                   break;
               }

               SortUtils.swap(a, i, j);
           }

           SortUtils.swap(a, lo, j); // Put v = a[j] into position
           return j; 
       }

   }
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1 回答 1

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您将推送和弹出与您的类型无关的整数T,因此您可能想要使用Stack<Integer>.

于 2017-03-10T07:40:10.230 回答