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
题目说 one pass ; 那么,这样,用两个直指针; p先跑n步,然后q和p一起跑,那么p跑到最后,q就正好在倒数第n个上面了。
Calculate the length first, and then remove the nth from the beginning.
public ListNode removeNthFromEnd(ListNode head, int n) { if(head == null) return null; //get length of list ListNode p = head; int len = 0; while(p != null){ len++; p = p.next; } //if remove first node int fromStart = len-n+1; if(fromStart==1) return head.next; //remove non-first node p = head; int i=0; while(p!=null){ i++; if(i==fromStart-1){ p.next = p.next.next; } p=p.next; } return head; }
Use fast and slow pointers. The fast pointer is n steps ahead of the slow pointer. When the fast reaches the end, the slow pointer points at the previous element of the target element.
public ListNode removeNthFromEnd(ListNode head, int n) { if(head == null) return null; ListNode fast = head; ListNode slow = head; for(int i=0; i<n; i++){ fast = fast.next; } //if remove the first node if(fast == null){ head = head.next; return head; } while(fast.next != null){ fast = fast.next; slow = slow.next; } slow.next = slow.next.next; return head; }
解题思路:加一个头结点dummy,并使用双指针p1和p2。p1先向前移动n个节点,然后p1和p2同时移动,当p1.next==None时,此时p2.next指的就是需要删除的节点。
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @return a ListNode def removeNthFromEnd(self, head, n): dummy=ListNode(0); dummy.next=head p1=p2=dummy for i in range(n): p1=p1.next while p1.next: p1=p1.next; p2=p2.next p2.next=p2.next.next return dummy.next