转载

92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

难度：medium

题目：反转从m到n的链表元素。一次遍历。

思路：记录m及m之前的位置，然后使用头插法。

Runtime: 2 ms, faster than 97.09% of Java online submissions for Reverse Linked List II.

Memory Usage: 36.9 MB, less than 0.95% of Java online submissions for Reverse Linked List II.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m == n) {
            return head;
        }
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode ptr = head, prevMPtr = dummyHead, tailPtr = head;
        for (int i = 1; i <= n; i++) {
            ListNode node = ptr;
            ptr = ptr.next;
            if (i == m - 1) {
                prevMPtr = node;
            } else if (i == m) {
                tailPtr = node;
                node.next = null;
            } else if (i > m) {
                node.next = prevMPtr.next;
                prevMPtr.next = node;
            }
        }
        tailPtr.next = ptr;
        
        return dummyHead.next;
    }
}

原文 https://segmentfault.com/a/1190000018144094

正文到此结束