Offer.P22[链表中倒数第k个节点]

方法一：模拟

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode p = head;
        int n = 0;
        while (p != null) {
            ++n;
            p = p.next;
        }
        p = head;
        for (int i = 1; i <= n - k; ++i) {
            p = p.next;
        }
        return p;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(1)$

方法二：双指针

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode slow = head, fast = head;
        while (k-- > 0) {
            fast = fast.next;
        }
        while (fast != null) {
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(1)$