[leetcode 19] Remove Nth Node From End of List

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，并返回修改后的链表头。该算法仅遍历一次链表，通过两个指针（快慢指针）定位待删除节点并进行操作。

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.

/**
 * 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) {
        if (!head || n==0) {
            return head;
        }
        ListNode dummy(-1);
        dummy.next = head;
        auto slow = &dummy;
        auto fast = head;
        for (int i = 0; i < n; i++) {
            fast = fast->next;
        }
        while (fast) {
            slow = slow->next;
            fast = fast->next;
        }
        auto tmp = slow->next;
        slow->next = slow->next->next;
        delete tmp;
        return dummy.next;
    }
};