"

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.

"

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
           Map<Integer, ListNode> map = new HashMap<Integer, ListNode>();
        ListNode t = new ListNode(0);
        t.next = head;
        int i = 1;
        while (t.next != null) {
            map.put(i, t.next);
            i++;
            t = t.next;
        }
        if (head.next == null) {
            return head.next;
        } else if (!map.containsKey(i - n - 1)) {
            head = map.get(i - n + 1);
        } else if (!map.containsKey(i - n + 1)) {
            map.get(i - n - 1).next = null;
        } else {
            map.get(i - n - 1).next = map.get(i - n + 1);
        }
        return head;
    }
}

链表注意考虑既是头又是尾，头，尾，中间等情况