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Array-1 CodingBat unlucky1 (java) 挑战

[英]Array-1 CodingBat unlucky1 (java) Challenge

I'm a senior in high school taking a computer science class. For homework, we have to create solutions to certain CodingBat (practice coding website) problems.我是一名计算机科学专业的高中生 class。对于家庭作业,我们必须为某些 CodingBat(实践编码网站)问题创建解决方案。 I am experiencing problems with this question, some of which include OutOfBounds for the array.我遇到了这个问题的问题,其中一些问题包括数组的 OutOfBounds。 Based on my code, I can't quite figure out why this is happening.根据我的代码,我不太明白为什么会这样。 The following attached code (below) is what I have created as a solution to the CodingBat problem for unlucky1 in Array-1 (java), which describes the challenge as: “We'll say that a 1 immediately followed by a 3 in an array is an "unlucky" 1. Return true if the given array contains an unlucky a in the first 2 or last 2 positions in the array.以下附加代码(下方)是我为 Array-1 (java) 中 unlucky1 的 CodingBat 问题创建的解决方案,它将挑战描述为:“我们会说 1 紧接着是 3 array 是一个“倒霉的”1。如果给定的数组在数组的前 2 个或最后 2 个位置包含倒霉的 a,则返回 true。

public boolean unlucky1(int[] nums) {
  int i = 0;
  for(i = 0; i < nums.length; i++)
    if(nums[i-1] == 1 && nums[i] == 3)
    {
      return true;
    }
    return false;
}

The problem statement is "Return true if the given array contains an unlucky a in the first 2 or last 2 positions in the array.", so you don't even need a loop - you just need to examine the first two and last two elements of the array:问题陈述是“如果给定数组在数组的前 2 个或后 2 个位置包含一个不幸的 a,则返回 true。”,因此您甚至不需要循环 - 您只需要检查前两个和后两个数组元素:

public boolean unlucky1(int[] nums) {
    return nums != null &&
           nums.length >= 2 &&
           (nums[0] == 1 && nums[1] == 3 ||
            nums[nums.length - 2] == 1 && nums[nums.length -1] == 3);
}

The following code is the correct method.下面的代码是正确的方法。

0.public static boolean unlucky1(int[] nums){  //Firstly,declare the method 
                                               "static"
    1.  int length = nums.length;//Get the array length.
 
    2.if((nums[0] == 1 && nums[1] == 3) && ( nums[length - 2] == 1 && 
                                             nums[length] -1  == 3)){
    3.      return true;        
                } 
    4.  return false;       
                }

In row 2,your code was: "if(nums[i-1] == 1 && nums[i] == 3)";在第 2 行,您的代码是:“if(nums[i-1] == 1 && nums[i] == 3)”;

It showed arrayoutofbound because the starting array index is 0 and you decalred in the if statement它显示 arrayoutofbound 因为起始数组索引为 0 并且您在 if 语句中贴标

" if(nums[0-1]...)which says if nums[-1] which is out of bounds."

Also to check the last 2 numbers of the array you do the following:还要检查数组的最后 2 个数字,您可以执行以下操作:

      ( nums[length - 1] == 1 && nums[length] == 3)) where : 

               " nums[length - 2] == 1" 
      checks 1 value before the last array value 
                         
                     and 
               " nums[length] - 1 == 3 "
             checks the last array value

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