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min(X, key=lambda x: x[-1])以使我更清楚（请参阅注释以获取详细信息）

Hoare algorithm

所以问题是：是否可以将数组传递给分区函数，所以下面的行会被执行？

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;
}

Answer 1

要执行它，leftIndex必须至少是rightIndex + 2，并且这是不可能发生的，假设我们用leftIndex＆lt; = rightIndex启动函数：

使用这两个循环：

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何一侧。

如果是这种情况，我们将离开该函数：

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），在执行之后它们也会在leftIndex == rightIndex + 1。我们仍然没有得到2的差异。

快速排序Hoare阵列分区

1 个答案: