java - 快速排序霍尔数组分区

Question

试图弄清楚为什么 Hoare 分区算法总是将数组分成两个正确的部分。在下面的代码中，我扩展了 Hoare algorithm 以使其对我来说更清楚(有关详细信息，请参阅评论)

int partition(int[] arr, int leftIndex, int rightIndex) {
  int pivot = arr[(leftIndex + rightIndex) / 2];

  while (leftIndex <= rightIndex) {
    while (arr[leftIndex] < pivot) leftIndex++;
    while (arr[rightIndex] > pivot) rightIndex--;

    // If all numbers at right places, than leftIndex and rightIndex 
    // could point at same array element index
    // So it's means partion done. 
    // We should return leftIndex + 1 cause 
    // rightIndex points at the last element of the left sub array

    if (leftIndex == rightIndex) return leftIndex + 1; 

    if (leftIndex < rightIndex) {
      swap(arr, leftIndex, rightIndex);
      leftIndex++;
      rightIndex--;
    }
  }

  //But here the tricky thing: Why does this "if case" never execute?
  if (leftIndex - 1 > rightIndex) 
    System.out.println("leftIndex - 1 > rightIndex");

  return leftIndex;
}

所以问题是:是否可以将数组传递给分区函数，以便执行下面的行？

if (leftIndex - 1 > rightIndex) 
  System.out.println("leftIndex - 1 > rightIndex");?

Answer 1

要执行此操作，leftIndex 必须至少为 rightIndex + 2，而这是不可能发生的，假设我们以 leftIndex <= rightIndex 启动函数:

用这两个循环:

while (arr[leftIndex] < pivot) leftIndex++;
while (arr[rightIndex] > pivot) rightIndex--;

指数永远不会相互交叉 - 如果不是更快，它们将停在枢轴的任一侧。

如果是这种情况，我们就离开这个函数:

if (leftIndex == rightIndex) return leftIndex + 1; 

所以，唯一剩下的就是:

if (leftIndex < rightIndex) {
  swap(arr, leftIndex, rightIndex);
  leftIndex++;
  rightIndex--;
}

即使它们尽可能接近(leftIndex == rightIndex - 1)，执行后它们也会在 leftIndex == rightIndex + 1 .我们仍然没有得到 2 的差异。