找到数组中相同数字之间的最大跨度 [英] Find the largest span between the same number in an array

问题描述

圣诞快乐，希望你精神振奋，我在Java-Arrays中有一个问题，如下图所示。我很挣扎着努力获得它。

Merry Christmas and hope you are in great Spirits,I have a Question in Java-Arrays as shown below.Im stuck up with this struggling to get it rite.

考虑数组中某些值的最左侧和最右侧的外观。我们会说跨度是两者之间的元素数量。单个值的跨度为1.编写一个** Java函数**，它返回给定数组中找到的最大跨度。

**示例：

maxSpan（{1,2,1,1,3}）→4，答案是4 coz MaxSpan介于1到1之间是4

maxSpan（{1 ，4,2,1,4,1,4}）→6，答案是6考斯MaxSpan 4到4之间是6

maxSpan（{1,4,2,1,4,4， 4}）→6，答案是6 coz Maxspan在4到4之间是6，大于MaxSpan在1和1之间是4，因此6> 4答案是6.

**Example:
maxSpan({1, 2, 1, 1, 3}) → 4,answer is 4 coz MaxSpan between 1 to 1 is 4
maxSpan({1, 4, 2, 1, 4, 1, 4}) → 6,answer is 6 coz MaxSpan between 4 to 4 is 6
maxSpan({1, 4, 2, 1, 4, 4, 4}) → 6,answer is 6 coz Maxspan between 4 to 4 is 6 which is greater than MaxSpan between 1 and 1 which is 4,Hence 6>4 answer is 6.

我的代码不起作用，它包含给定元素的所有Spans，我无法找到给定元素的MaxSpan。

I have the code which is not working,it includes all the Spans for a given element,im unable to find the MaxSpan for a given element.

请帮助我出去了。

上述计划的结果如下所示

预计此次运行

maxSpan（{1,2,1,1,3}）→4 5 X

maxSpan（{1,4,2,1,4， 1,4}}→6 8 X

maxSpan（{1,4,2,1,4,4,4}）→6 9 X

maxSpan（{3， 3,3}）→3 5 X

maxSpan（{3,9,3}）→3 3 OK

maxSpan（{3,9,9}）→2 3 X

maxSpan（{3 ，9}）→1 1 OK

maxSpan（{3,3}）→2 3 X

maxSpan（{}）→0 1 X

maxSpan（{1}）→1 1 OK

:: Code ::

::Code::

public int maxSpan(int[] nums) {    
    int count=1;//keep an intial count of maxspan=1    
    int maxspan=0;//initialize maxspan=0    
    for(int i=0;i<nums.length;i++){    
        for(int j=i+1;j<nums.length;j++){    
              if(nums[i] == nums[j]){
                 //check to see if "i" index contents == "j" index contents    
                 count++;    //increment count
                 maxspan=count;  //make maxspan as your final count  
                 int number = nums[i]; //number=actual number for maxspan               
              }                            
        }       
    }    
  return maxspan+1; //return maxspan        
}    

推荐答案

我明白了您尝试的以下问题：

I see the following problems with your attempt:

• 您的计数完全错误。您可以从 i 和 j 中计算 count ： j - i + 1

• Your count is completely wrong. You can instead calculate count from i and j: j - i + 1

你要重写 maxcount 只要您获得任何范围，那么您将最终得到最后范围，而不是最大范围。通过 maxspan = Math.max（maxspan，count）; 。

You're overriding maxcount as soon as you get any span, so you're going to end up with the last span, not the maximum span. Fix it by going maxspan = Math.max(maxspan, count);.

修复它你可以删除它行 int number = nums [i]; ，因为你从不使用数字。

You can remove the line int number = nums[i]; as you never use number.

在返回 maxspan + 1中删除 +1 ;`如果你按照上面的提示。

Remove the +1 in the returnmaxspan+1;` if you follow my tips above.

如果数组中有任何值，则初始 maxspan 应该为1如果数组为空，则为0。

Initial maxspan should be 1 if there are any values in the array, but 0 if the array is empty.

这应该可以帮助您使其正常工作。请注意，您可以在数组的一次传递中执行此操作，但这可能会让您感觉太过分了。在考虑效率之前，集中精力让代码工作。

That should help you get it working. Note that you can do this in a single pass of the array, but that's probably stretching it too far for you. Concentrate on getting your code to work before considering efficiency.

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