題目
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
思路
這個問題的核心就在於只遍歷一次,如果允許兩次則很好解決、先第一遍遍歷知道鏈表長度、第二遍就直接除去倒數第N個結點了。
只遍歷一次我們就想,遍歷結束的標誌就是指向爲空、從指向爲空的結點往前數N個點即爲目標結點,也就是說目標結點與結束標誌相差爲N,那麼我們就可以設置兩個指針p和q,中間相差爲N,兩個指針同步向後走,那麼p(假設在前面)指向爲空的時候,那麼q就自然爲目標指針了,剩下就完成刪除結點的操作。
代碼
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode p = new ListNode(0);
ListNode q = p;
ListNode result = p;
p.next = head;
for(int i = 0 ; i <= n ; ++ i){
p = p.next;
}
while(p != null){
p = p.next;
q = q.next;
}
q.next = q.next.next;
return result.next;
}
}