Leetcode19. Remove Nth Node From End of List

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本文详细解析了LeetCode第19题“从链表末尾删除第n个节点”的解决方案，通过双指针技巧，仅用一次遍历即可完成任务。提供了Java、C++及Python三种语言的实现代码。

Leetcode19. Remove Nth Node From End of List

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:
Could you do this in one pass?

解析

设想假设设定了双指针 p 和q 的话，当 q 指向末尾的 NULL，p 与 q 之间相隔的元素个数为 n 时，那么删除掉p 的下一个指针就完成了要求。

设置虚拟节点 dummyHead 指向 head
设定双指针 p 和 q，初始都指向虚拟节点dummyHead
移动 q，直到 p 与 q 之间相隔的元素个数为 n
同时移动 p 与 q，直到q指向的为 NULL
将 p 的下一个节点指向下下个节点

Java

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
}

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
     ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;

        ListNode* p = dummyHead;
        ListNode* q = dummyHead;
        for( int i = 0 ; i < n + 1 ; i ++ ){
            q = q->next;
        }

        while(q){
            p = p->next;
            q = q->next;
        }

        ListNode* delNode = p->next;
        p->next = delNode->next;
        delete delNode;
        
		//dummyHead被设置成了0，如果直接返回dummyHead，会在首部打印多余的0。
        ListNode* retNode = dummyHead->next;
        delete dummyHead;

        return retNode;
        
    }
};

Python

def removeNthFromEnd(self, head, n):
    dummy = ListNode(0)
    dummy.next = head
    fast = slow = dummy
    for _ in xrange(n):
        fast = fast.next
    while fast and fast.next:
        fast = fast.next
        slow = slow.next
    slow.next = slow.next.next
    return dummy.next