19. Remove Nth Node From End of List

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