（3-2）单链表面试题

来源：https://www.bilibili.com/video/BV1E4411H73v?from=search&seid=150214088132003010&spm_id_from=333.337.0.0

反转单链表

力扣解析：https://leetcode-cn.com/problems/reverse-linked-list/solution/shi-pin-jiang-jie-die-dai-he-di-gui-hen-hswxy/

public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode curr = head;
    while (curr != null) {
        ListNode next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

作者：tangweiqun
链接：https://leetcode-cn.com/problems/reverse-linked-list/solution/shi-pin-jiang-jie-die-dai-he-di-gui-hen-hswxy/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

从尾到头打印单链表

辅助栈法

class Solution {
    public int[] reversePrint(ListNode head) {
        LinkedList<Integer> stack = new LinkedList<Integer>();
        while(head != null) {
            stack.addLast(head.val);
            head = head.next;
        }
        int[] res = new int[stack.size()];
        for(int i = 0; i < res.length; i++)
            res[i] = stack.removeLast();
    return res;
    }
}

作者：jyd
链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/solution/mian-shi-ti-06-cong-wei-dao-tou-da-yin-lian-biao-d/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

合并两个有序链表

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        else if (l2 == null) {
            return l1;
        }
        else if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }
        else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }

    }
}

作者：z1m
链接：https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/yi-kan-jiu-hui-yi-xie-jiu-fei-xiang-jie-di-gui-by-/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

理解：