删除链表倒数第n个节点

删除链表倒数第n个节点

解法一:
直接遍历出链表长度L，然后删除第L-n+1个节点即可：

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0, head);
        int length = getLength(head);
        ListNode cur = dummy;
        for (int i = 1; i < length - n + 1; ++i) {
            cur = cur.next;
        }
        cur.next = cur.next.next;
        ListNode ans = dummy.next;
        return ans;
    }

    public int getLength(ListNode head) {
        int length = 0;
        while (head != null) {
            ++length;
            head = head.next;
        }
        return length;
    }
}

解法二：
通过递归回溯遍历：

    public ListNode removeNthFromEnd(ListNode head, int n) {
    	int traverse = traverse(head, n);
    	if(traverse == n)
    	    return head.next;
    	return head;
    }
    
    private int traverse(ListNode node, int n) {
        if(node == null)
            return 0;
        int num = traverse(node.next, n);
        if(num == n)
            node.next = node.next.next;
        return num + 1;
    }

收藏整理自LeetCode.