在java中比较相同数组的元素 [英] comparing elements of the same array in java

问题描述

我试图比较相同数组的元素。这意味着我想比较0元素与每个其他元素，1元素与每个其他元素等。问题是，它不能按预期工作。 。我做的是我有两个for循环从0到array.length-1 ..然后我有一个if语句如下：if（a [i]！= a [j + 1]）

  for（int i = 0; i  for（int k = 0 ; k  if（a [i]！= a [k + 1]）{
 System.out.println（a [i] +not与+ a [k + 1] +\\\
）相同; 
} 
} 
} 
  

解决方案>

首先，你需要循环到< a.length ，而不是 a.length - 1 。因为这是严格小于你需要包括上限。

所以，检查所有对元素你可以做：

  for（int i = 0; i  for（int k = 0; k  if（a [i]！= a [k]）{
 // do stuff 
} 
} 
} 
  

但这会比较，例如 a [2] c $ c> a [3] ，然后 a [3] 到 a [2] 。鉴于您正在检查！= 这似乎是浪费。

更好的方法是比较每个元素<$

 到  for（int i = 0; i  for（int k = i + 1; k  if（a [i] ！= a [k]）{
 // do stuff 
} 
} 
} 
  

因此，如果你有索引[1 ... 5]，比较将进行



1. code> 1 - > 2


2. 1 - > 3


3. 1 - > 4


4. 1 - > 5


5. 2 - > 3


6. 2 - > 4


7. 2 - > 5


8. 3 - > 4


9. 3 - > 5


10. 4 - > 5

因此，您看不到重复的对。想想一个人的圈子，所有人都需要彼此握手。

I am trying to compare elements of the same array. That means that i want to compare the 0 element with every other element, the 1 element with every other element and so on. The problem is that it is not working as intended. . What i do is I have two for loops that go from 0 to array.length-1.. Then i have an if statement that goes as follows: if(a[i]!=a[j+1])

for (int i = 0; i < a.length - 1; i++) {
    for (int k = 0; k < a.length - 1; k++) {
        if (a[i] != a[k + 1]) {
            System.out.println(a[i] + " not the same with  " + a[k + 1] + "\n");
        }
    }
}

解决方案

First things first, you need to loop to < a.length rather than a.length - 1. As this is strictly less than you need to include the upper bound.

So, to check all pairs of elements you can do:

for (int i = 0; i < a.length; i++) {
    for (int k = 0; k < a.length; k++) {
        if (a[i] != a[k]) {
            //do stuff
        }
    }
}

But this will compare, for example a[2] to a[3] and then a[3] to a[2]. Given that you are checking != this seems wasteful.

A better approach would be to compare each element i to the rest of the array:

for (int i = 0; i < a.length; i++) {
    for (int k = i + 1; k < a.length; k++) {
        if (a[i] != a[k]) {
            //do stuff
        }
    }
}

So if you have the indices [1...5] the comparison would go

1. 1 -> 2
2. 1 -> 3
3. 1 -> 4
4. 1 -> 5
5. 2 -> 3
6. 2 -> 4
7. 2 -> 5
8. 3 -> 4
9. 3 -> 5
10. 4 -> 5

So you see pairs aren't repeated. Think of a circle of people all needing to shake hands with each other.

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