因此,我尝试在 Java 中构建一个方法,该方法在调用时将作为参数将反转 LinkedList
末尾的元素数。
所以,如果我有
{hat cat bat mat}
然后我输入2作为我的参数,那么最后2个元素就会这样反转
{hat cat mat bat}
这是我尝试过的:
public void reverseLastFew(int howMany)
{
int s = size();
LinkedListIterator iter1 = new LinkedListIterator();
LinkedListIterator iter2 = new LinkedListIterator();
for (int i=0;i<s-howMany;i++)
{
iter1.next();
}
for (int i = 0;i<s;i++)
{
iter2.next();
}
Object temp = null;
while (iter2.hasNext())
{
temp = iter2.next();
}
iter2.remove();
iter1.add(temp);
}
最佳答案
我有一个类似问题的解决方案:
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given "1"->"2"->"3"->"4"->"5"->NULL, m = 2 and n = 4,
return "1"->"4"->"3"->"2"->"5"->NULL.
解决方法:
/**
* Definition for singly-linked list.
* public class ListNode {
* String val;
* ListNode next;
* ListNode(String x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if (head==null) {
return null;
}
ListNode dummy = new ListNode(" ");
dummy.next = head;
head = dummy;
for (int i=1; i<m; i++) {
head = head.next;
}
ListNode pre = head;
ListNode start = head.next;
ListNode end = start;
ListNode post = start.next;
for (int i=m; i<n; i++) {
if (post==null) {
return null;
}
ListNode temp = post.next;
post.next = end;
end = post;
post = temp;
}
start.next = post;
pre.next = end;
return dummy.next;
}
}
因此,您可以根据已有的计算m
和n
,或者修改此解决方案以直接解决您的问题。无论如何,这种就地一次性解决方案非常好。
关于java - 如何反转这个从第 n 个元素到最后的链表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36489501/