Question

在Codility上，存在一个问题，即计算青蛙到达Y位置所需的最小跳数。

问题如下：

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:

  X = 10
  Y = 85
  D = 30
the function should return 3, because the frog will be positioned as follows:

after the first jump, at position 10 + 30 = 40
after the second jump, at position 10 + 30 + 30 = 70
after the third jump, at position 10 + 30 + 30 + 30 = 100

我已经“解决”了问题，并得到“ 3”的结果，就像在示例中一样。但是，当我提交代码时，除示例外，我只得到11％的成绩，并且所有测试均未通过。

这是我的代码

int count = 0;

while(X <= Y){
X += D;
count++;
}       

return count;

在我的代码中，我基本上算出到达Y所需的跳跃，这是青蛙想要到达的位置。我没有正确理解问题？如果是这样，我想念的是什么？

Answer 1

您正在使用条件

while(X <= Y)

进行新的跳转。但是，在X == Y的情况下，您已经达到了目标，并且不想进一步前进。因此，更新为

while(X < Y) //as long as we have not reached the goal

Answer 2

这是我的解决方案。您的距离计算为Y-X。因此，您只需要划分并检查除数即可：

    function solution(X,Y,D) {

    var distance = Y-X;
    var steps = distance/D;
    /* If steps number is rounded integer just return steps */
    if((steps*D) == distance && Number.isInteger(steps)) {
      return steps;
    } else {
/* if there is rest of division then just round it (to floor) and increase steps number for one */
      return Math.floor(steps)+1;
    }

    }

    console.log(solution(10,85,30)); /* 3 */

这是JavaScript的解决方案，复杂度基本上是O（1）。

Answer 3

为什么在这里使用任何循环，循环是一件昂贵的事情。只需一行答案

function solution(X,Y,D) {
    return Math.ceil((Y-X)/D);
}

如何修复“ Codility FrogJump”算法？

3 个答案: