英文:
removeRange method with ArrayList
问题
以下是您要翻译的内容：
对于课程中的这个方法，我遇到了困难。我必须编写一个方法，该方法接受最大值和最小值的范围，并删除所有包含在这些值之间的元素。例如，如果我在主方法中调用removeRange(list, 5, 7);，它应该将以下值：[7, 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7]，更改为：[9, 4, 2, 3, 1, 8]。但是，当我编译时，我始终得到[9, 4, 2, 7, 3, 1, 8, 7]。以下是带有参数的removeRange方法。
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for (int i = 0; i < list.size(); i++) {
if ((list.get(i) >= min && list.get(i) <= max)) {
list.remove(i);
}
}
}
英文:
Having trouble figuring this method out for class. I have to write a method that takes the range of a max and min and removes all elements containing those values. FOr example, if I put in the main method removeRange(list, 5, 7); in the main method, it should take the values of [7, 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7] and turn it to [9, 4, 2, 3, 1, 8] . But I keep getting [9, 4, 2, 7, 3, 1, 8, 7] when I compile. Here is my removeRange method with the parameters.
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for(int i = 0; i < list.size(); i++) {
if((list.get(i) >= min && list.get(i) <= max)) {
list.remove(i);
}
}
}
答案1
得分: 2
当您从您的 ArrayList 中移除一个元素时，随之移除的元素后面的索引会减少，因此在循环的下一次迭代中，您的循环会跳过下一个元素。
您可以通过在移除元素时将 i 减少来进行调整：
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for(int i = 0; i < list.size(); i++) {
if((list.get(i) >= min && list.get(i) <= max)) {
list.remove(i);
i--;
}
}
}
或者通过反向迭代：
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for(int i = list.size() - 1; i >= 0; i--) {
if((list.get(i) >= min && list.get(i) <= max)) {
list.remove(i);
}
}
}
英文:
When you remove an element from your ArrayList, the indices of the elements following the removed element are decremented, so your loop would skip the next element in the following iteration.
You can account for it by decrementing i when you remove an element:
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for(int i = 0; i < list.size(); i++) {
if((list.get(i) >= min && list.get(i) <= max)) {
list.remove(i);
i--;
}
}
}
Or by iterating backwards:
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for(int i = list.size() - 1; i >= 0; i--) {
if((list.get(i) >= min && list.get(i) <= max)) {
list.remove(i);
}
}
}
答案2
得分: 2
在你的数组 7, 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7 中，迭代如下 ->
为了方便，让范围为7到7。
最初，i = 0，大小 = 14：移除了7。大小 = 13
新数组 - 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7
现在 i = 1，大小 = 13：您现在正在访问a[1]，即4，所以跳过了9。
稍后 i = 3，大小 = 13，移除了7。大小 = 12
新数组 - 9, 4, 2, 7, 5, 3, 5, 1, 7, 8, 6, 7
现在 i = 4，大小 = 12：您现在正在访问a[4]，即5，所以跳过了7。
希望你现在能看到你的逻辑中的缺陷。
尝试每次移除一个元素时执行i = i-1。
英文:
In your array 7, 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7 the iteration goes like this ->
Let range be 7 to 7 for ease.
initially, i = 0, size = 14: 7 is removed. size = 13
new array - 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7
now i = 1, size = 13: you are now accessing a[1] ie, 4 so 9 is skipped.
later i = 3, size =13, 7 is removed. size = 12
new array - 9, 4, 2, 7, 5, 3, 5, 1, 7, 8, 6, 7
now i = 4, size = 12: you are now accessing a[4] ie, 5 so 7 is skipped.
hope the flaw in your logic is visible to you now.
Try doing i = i-1 every time you remove an element.
答案3
得分: 0
你可以像这样使用 Iterator。
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for (Iterator<Integer> i = list.iterator(); i.hasNext();) {
int element = i.next();
if (element >= min && element <= max)
i.remove();
}
}
以及
ArrayList<Integer> list = new ArrayList<>(List.of(7, 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7));
removeRange(list, 5, 7);
System.out.println(list);
输出：
[9, 4, 2, 3, 1, 8]
英文:
You can use Iterator like this.
public static void removeRange(ArrayList<Integer> list, int min, int max) {
for (Iterator<Integer> i = list.iterator(); i.hasNext();) {
int element = i.next();
if (element >= min && element <= max)
i.remove();
}
}
and
ArrayList<Integer> list = new ArrayList<>(List.of(7, 9, 4, 2, 7, 7, 5, 3, 5, 1, 7, 8, 6, 7));
removeRange(list, 5, 7);
System.out.println(list);
output:
[9, 4, 2, 3, 1, 8]
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