[刷题]Remove Nth Node From End of List

原创于 2015-04-29 15:05:12 发布 · 329 阅读

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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。该算法仅需遍历一次链表，通过双指针技巧实现目标节点的定位与删除操作。

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[LintCode]Remove Nth Node From End of List

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: The head of linked list.
     */
    ListNode removeNthFromEnd(ListNode head, int n) {
        // 2015-04-29 O(n)
        if (head == null) {
            return null;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head; //被删除的节点可能是头节点，所以要用dummy
        head = dummy;
        ListNode nthNode = dummy;
        for (int i = 0; i < n; i++) {
            nthNode = nthNode.next;
            if (nthNode == null) {
                return null;
            }
        }
        while (nthNode.next != null) {
            head = head.next;
            nthNode = nthNode.next;
        }
        // 删掉head的下一个节点
        head.next = head.next.next;
        return dummy.next;
    }
}