[LeetCode] 19. Remove Nth Node From End of List 移除链表倒数第N个节点
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.
移除给定链表的倒数第n个节点，n总是有效的，要求一次遍历完成one pass。
解法1: two pass，先求出链表的总长度，然后在删除node
解法2：双指针，经典题。主要是在一个遍历中找到第n个倒数元素，一个指针先走n步，然后两个指针同步走，直到第一个走到终点，第二个指针指向的就是需要删除的节点。Corner case: head == null; 头节点的处理，比如，1->2->NULL, n =2; 这时，要删除的就是头节点。
Java: Naive Two Pass
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;
}
Java:
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;
}
Python:
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def __repr__(self):
if self is None:
return "Nil"
else:
return "{} -> {}".format(self.val, repr(self.next))
class Solution:
# @return a ListNode
def removeNthFromEnd(self, head, n):
dummy = ListNode(-1)
dummy.next = head
slow, fast = dummy, dummy
for i in xrange(n):
fast = fast.next
while fast.next:
slow, fast = slow.next, fast.next
slow.next = slow.next.next
return dummy.next
C++:
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if (!head->next) return NULL;
ListNode *pre = head, *cur = head;
for (int i = 0; i < n; ++i) cur = cur->next;
if (!cur) return head->next;
while (cur->next) {
cur = cur->next;
pre = pre->next;
}
pre->next = pre->next->next;
return head;
}
};
