Question

我做了一个遍历数组的函数，并打印出该数组的任何两个值，它们的总和为K。外部for循环为O（n），但如果运行时是O（Log n）或O（n）。你能帮忙吗？谢谢！！

int canMakeSum(int *array, int n, int key){
  int i, j;

  for(i = 0; i < n; i++){
    for(j = (i+1); j < n; j++){
      if(array[i]+array[j] == key){
        printf("%d + %d = %d\n", array[i], array[j], key);
      }
    }
  }
}

Answer 1

由于内部循环取决于外部循环的值，因此如果不对两者进行分析，就无法找到总程序图的复杂性。内部循环的复杂度为n - i -1。

如果要计算程序的复杂度，可以将i = 0从i = n - 1到T(n) = (n - 1) + (n-2) + ... + 1 + 0 = (n-1)n/2 = \Theta(n^2)求和。因此，总复杂度为\Theta(1)（因为内部循环中的语句具有恒定的复杂度（public class AugmentedImageNode extends AnchorNode { private static final String TAG = "AugmentedImageNode"; // The augmented image represented by this node. private AugmentedImage image; // Models of the 4 corners. We use completable futures here to simplify // the error handling and asynchronous loading. The loading is started with the // first construction of an instance, and then used when the image is set. private static CompletableFuture<Void> rendobject; private ViewRenderable testViewRenderable; //ImageView imgView; public AugmentedImageNode(Context context) { // Upon construction, start loading the models for the corners of the frame. if (rendobject == null) { ImageView imgView = imgView.inflate(context, R.layout.imgboard, null); Picasso.get() .load("http://scoopak.com/wp-content/uploads/2013/06/free-hd-natural-wallpapers-download-for-pc.jpg") .into(imgView); rendobject = ViewRenderable.builder() .setView(context, imgView) .setVerticalAlignment(ViewRenderable.VerticalAlignment.BOTTOM) .setSizer(new FixedHeightViewSizer(0.12f)) .build() .thenAccept(renderable -> { testViewRenderable = renderable; //testViewRenderable = renderable; }); /* ModelRenderable.builder() .setSource(context, Uri.parse("models/tinker.sfb")) .build(); */ } } /** * Called when the AugmentedImage is detected and should be rendered. A Sceneform node tree is * created based on an Anchor created from the image. The cornerNode is then positioned based on the * extent of the image. There is no need to worry about world coordinates since everything is * relative to the center of the image, which is the parent node of the corner. */ @SuppressWarnings({"AndroidApiChecker", "FutureReturnValueIgnored"}) public void setImage(AugmentedImage image) { this.image = image; // If any of the models are not loaded, then recurse when all are loaded. if (!rendobject.isDone()) { CompletableFuture.allOf(rendobject)//, urCorner, llCorner, lrCorner) .thenAccept((Void aVoid) -> setImage(image)) .exceptionally( throwable -> { Log.e(TAG, "Exception loading", throwable); return null; }); } // Set the anchor based on the center of the image. setAnchor(image.createAnchor(image.getCenterPose())); // Make the node(s). Vector3 localPosition = new Vector3(); Node cornerNode; //localPosition.set(-0.5f * image.getExtentX(), 0.0f, -0.5f * image.getExtentZ()); localPosition.set(-0.0f * image.getExtentX(), 0.1f, +0.5f * image.getExtentZ()); cornerNode = new Node(); cornerNode.setParent(this); cornerNode.setLocalPosition(localPosition); cornerNode.setLocalRotation(Quaternion.axisAngle(new Vector3(-1f, 0, 0), 90f)); //cornerNode.setLocalScale(scaledWidth, scaledHeight, scaledWidth); cornerNode.setRenderable(testViewRenderable); } }）。

Answer 2

正如其他人已经表明的那样，内循环仍然是 O（n）；这是n / 2次迭代的平均值，值1到n在外循环的迭代中平均分布。

是的，您可以在 O（n log n）中解决问题。

首先，对数组进行排序；这是n log n。现在，您有了一个线性（ O（n））过程来查找所有组合。

lo = 0
hi = n-1

while lo < hi {
   sum = array[lo] + array[hi]
   if sum == k {
       print "Success", array[lo], array[hi]
       lo += 1
       hi -= 1
   }
   else if sum < k        // need to increase total
       lo += 1
   else                   // need to decrease total
       hi -= 1

Answer 3

尽管内循环减少了外循环中每次迭代扫描的项目数量，但仍为O（n）。总时间复杂度为O（n ^ 2）。

假设您有25000个元素的数组。在i = 0和j = 1的起始点，j迭代通过的元素数（最坏的情况是没有与key匹配的元素）是24999个元素。与元素总数相比，差异很小，因此就像“通过” n个元素一样。

内循环的运行时间是多少？

3 个答案: