Question

在我的函数中，我必须从给定数组中找到给定数字K的所有唯一三元组。它会找到所有三元组，但其中很多都有两倍或更多，例如1 5 1与5 1 1或int triplet(int *array, int size, int K) { int i, j, k; int found = 0; /* triplets whose sum is equal to K */ for(i = 0; i < size; i++) { for (j = 0; j < size; j++) { for (k = 0; k < size; k++) { if(array[i] + array[j] + array[k] == K) { printf("Triplet Found : %d, %d, %d\n", array[i], array[j], array[k]); found++; } } } } return found; }等等相同。

有人可以帮我吗？

//Class that detects the collision
if (other.gameObject.tag == "enemy") 
{
    EnemyMovement enemyMove = other.GetComponent <EnemyMovement> ();
    if (enemyMove.slowMove != 1) {
        return;
    }

    enemyMove.Slow (2, 0.5f, true);

    //....

//Class that handles the Enemy-Movement
//The slowfactor gets multiplied with the Vector3 that handles the movementspeed of the Character

void FixedUpdate () 
{
    Movement();
}

void Movement()
{
    gegnerRigid.MovePosition (transform.position + (gegnerMove * slowMove));
}


public void Slow (float duration, float slowFactor, bool slowed)
{
    if (slowed) 
    {
        if (duration > 0) {
            duration -= Time.deltaTime;
            slowMove = slowFactor;
            Slow (duration, slowFactor, slowed);  //this recursive Call leads to huge performance issues, but how to keep the function going?
        } else {
            slowMove = 1;
            slowed = false;
        }
    }
}

Answer 1

要排除组合，您可以尝试以下更改：

int triplet(int *array, int size, int K) {
  int i, j, k;
  int found = 0; /* triplets whose sum is equal to K */
  for(i = 0; i < size; i++) {
    for (j = i; j < size; j++) {    
      for (k = j; k < size; k++) {
        if(array[i] + array[j] + array[k] == K) {
          printf("Triplet Found : %d, %d, %d\n", array[i], array[j], array[k]);
          found++;
        }           
      }
    }
  }
  return found;
}

请注意j = i声明中的k = j和for loop语句。这样，您将跳过针对您的案例的可能重复项，但涵盖i，j和k变量的所有变体。

Answer 2

您不应重复已处理的组合;一个可能的解决方案是从前一个ID + 1开始每个循环，如下所示：

for(i = 0; i < size; i++) {
 for (j = i+1; j < size; j++) {    
  for (k = j+1; k < size; k++) {
    if(array[i] + array[j] + array[k] == K) {
      printf("Triplet Found : %d, %d, %d\n", array[i], array[j], array[k]);
      found++;
    }           
  }
}

设置j = i + 1;可防止同一元素三元组！

寻找独特的三胞胎

2 个答案: