[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 n^th 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;
    }
};

　　

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