将ArrayList中特定的元素移动到最后一个索引位置。

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英文:

Move a specific element in arraylist to the last index

问题

假设我有这个列表

List<Integer> grades = Arrays.asList(9, 8, 1, 0, 7, 0, 0, 3);

我想将所有的零移动到最后的索引位置。所以我希望输出类似于:

9, 8, 1, 7, 3, 0, 0, 0

我可以在简单数组中做到这一点,但是我仍然没有找出如何在ArrayList中实现这一点。

英文:

Let's say I have this List

List&lt;Integer&gt; grades = Arrays.asList(9,8,1,0,7,0,0,3);

I want to move all zeros to the last indexes .. So I want the output to be something like:

9,8,1,7,3,0,0,0

I can do that with simple arrays but I still didn't figure out how to do that in ArrayList.

答案1

得分: 3

你可以使用Comparator.comparing对列表进行相应排序,将所有的 0 移动到末尾,而不对其余的值进行排序。

grades.sort(Comparator.comparing(value -&gt; value == 0 ? 1 /* 排在末尾 */ : 0 /* 不排序 */));  //[9, 8, 1, 7, 3, 0, 0, 0]
英文:

You can sort the list accordingly using Comparator.comparing, to move all 0 to last and not to sort the remaining values

grades.sort(Comparator.comparing(value -&gt; value == 0 ? 1 /* sort last */ : 0 /* don&#39;t sort */));  //[9, 8, 1, 7, 3, 0, 0, 0]

答案2

得分: 2

正如评论中所指出的,您可以通过使用Listgetset方法来重用您已经为数组编写的相同代码,而不是使用[]运算符。

另一种“奇特”的方法可能是利用排序保持两个等效元素的相对位置的特性,并使用自定义的Comparator对列表进行排序,该Comparator根据项目是否等于0进行排序 - 所有的零将被移到末尾,而所有其他项目将保持其相对位置:

grades.sort(Comparator.comparing(i -> i == 0));
英文:

As noted in the comments, you could reuse the same code you have for arrays by using the List's get and set methods instead of the [] operator.

Another "sneaky" approach could be to utilize the fact that sorting keeps the relative positions of two elements that are equivalent, and sort the list with a custom Comparator that sorts according to whether or not an item is equal to 0 - all the zeros will moved to the end, and all the other items will keep their relative postions:

grades.sort(Comparator.comparing(i -&gt; i == 0));

答案3

得分: 1

对于比使用 O(n log(n))sort() 解决方案更快的 O(n) 实现,可以按照以下方式对数组进行操作:

int j = 0;
for (Integer value : grades) {
    if (value != 0) {
        grades.set(j++, value);
    }
}
while (j < grades.size()) {
    grades.set(j++, 0);
}

当然,如果不是 ArrayList,使用 set(int, ?) 可能会影响性能,因此最好使用可更新的迭代器,即 ListIterator

ListIterator<Integer> iter = grades.listIterator();
for (Integer value : grades) {
    if (value != 0) {
        iter.next(); // 无需先调用 hasNext()
        iter.set(value);
    }
}
while (iter.hasNext()) {
    iter.next();
    iter.set(0);
}

作为对比,这是相应的数组解决方案:

int[] grades = {9,8,1,0,7,0,0,3};

int j = 0;
for (int i = 0; i < grades.length; i++) {
    if (grades[i] != 0) {
        grades[j++] = grades[i];
    }
}
while (j < grades.length) {
    grades[j++] = 0;
}
英文:

For a faster O(n) implementation than using an O(n log(n)) sort() solution, do it the same way you'd do it for an array:

int j = 0;
for (Integer value : grades) {
	if (value != 0) {
		grades.set(j++, value);
	}
}
while (j &lt; grades.size()) {
	grades.set(j++, 0);
}

Of course, using set(int, ?) can be bad for performance if not an ArrayList, so it'd be better to use an updatable iterator, i.e. a ListIterator:

ListIterator&lt;Integer&gt; iter = grades.listIterator();
for (Integer value : grades) {
	if (value != 0) {
		iter.next(); // No need to call hasNext() first
		iter.set(value);
	}
}
while (iter.hasNext()) {
	iter.next();
	iter.set(0);
}

For comparison, here is the comparable array solution:

int[] grades = {9,8,1,0,7,0,0,3};

int j = 0;
for (int i = 0; i &lt; grades.length; i++) {
	if (grades[i] != 0) {
		grades[j++] = grades[i];
	}
}
while (j &lt; grades.length) {
	grades[j++] = 0;
}

答案4

得分: 0

感谢大家。。

我成功地用和处理简单数组相同的方式做到了:


List<Integer> price = Arrays.asList(9,8,1,0,7,0,0,3);
       
int count = 0;

for(int i=0; i<price.size(); i++) {
    if(price.get(i) != 0) {
        price.set(count++, price.get(i));	
    }
}

while(count < price.size()) {
    price.set(count++, 0);
}

System.out.print(Arrays.toString(price.toArray()));

}
英文:

Thanks guys ..

I managed to do it the same way I did it with simple arrays :


 List&lt;Integer&gt; price = Arrays.asList(9,8,1,0,7,0,0,3);
       
        
        int count = 0;
        
        
        for(int i=0; i&lt;price.size(); i++) {
        	if(price.get(i) !=0) {
        		
        		price.set(count++, price.get(i));	
        		
        	}
        	
        }
        	 while(count&lt;price.size()) {
             	price.set(count++, 0);
             
             	
             }
        	System.out.print(Arrays.toString(price.toArray()));
        }

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  • 本文由 发表于 2020年9月19日 00:15:35
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