可以使这种算法更好吗? [英] Could this algorithm be made better?

问题描述

我有一个问题，在我写的一种算法中，要分离出左边的所有偶数和右边的奇数.

I have a question, in one of the algorithms, I had written, to seperate out, all the even numbers to the left and odd numbers to the right.

示例:输入:

{12，10，52，57，14，91，34，100，245，78，91，32，354，80，13，67，65}

{ 12, 10, 52, 57, 14, 91, 34, 100, 245, 78, 91, 32, 354, 80, 13, 67, 65 }

输出:

{12,10,52,80,14,354,34,100,32,78,91,245,91,57,13,67,65}

{12,10,52,80,14,354,34,100,32,78,91,245,91,57,13,67,65}

下面是算法

public int[] sortEvenAndOdd(int[] combinedArray) {
  int endPointer = combinedArray.length - 1;
  int startPointer = 0;
  for (int i = endPointer; i >= startPointer; i--) {
    if (combinedArray[i] % 2 == 0) {
      if (combinedArray[startPointer] % 2 != 0) {
        int temp = combinedArray[i];
        combinedArray[i] = combinedArray[startPointer];
        combinedArray[startPointer] = temp;
        startPointer = startPointer + 1;
      } else {
        while (combinedArray[startPointer] % 2 == 0 &&
          (startPointer != i)) {
          startPointer = startPointer + 1;
        }
        int temp = combinedArray[i];
        combinedArray[i] = combinedArray[startPointer];
        combinedArray[startPointer] = temp;
        startPointer = startPointer + 1;
      }
    }
  }
  return combinedArray;
}

任何人，有什么建议，因为它可以达到O(n)或更好?

Anybody, have any suggestions, for it make it to O(n) or better ?

推荐答案

您的代码为O(n)，但比所需的要复杂一些.这是一个改进.

Your code is O(n), but it's a bit more complicated than it needs to be. Here's an improvement.

startPointer = 0;
endPointer = a.length - 1;
while (startPointer < endPointer)
{
    if (a[startPointer] % 2 != 0)
    {
        // move endPointer backwards to even number
        while (endPointer > startPointer && a[endPointer] % 2 != 0)
        {
            --endPointer;
        }
        swap(a[startPointer], a[endPointer]);
    }
    ++startPointer;
}

顺便说一句，该操作更多的是分区而不是排序.我认为更好的函数名称是 partitionEvenOdd .

By the way, the operation is more of a partition than a sort. I think a better function name would be partitionEvenOdd.

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