链表相关算法

Leetcode 21. 合并两个有序链表

    /**
     * 解法一：迭代法
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }

        ListNode head;
        if (l1.val <= l2.val) {
            head = l1;
            l1 = l1.next;
        } else {
            head = l2;
            l2 = l2.next;
        }
        ListNode listNode = head;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                listNode.next = l1;
                l1 = l1.next;
            } else {
                listNode.next = l2;
                l2 = l2.next;
            }
            listNode = listNode.next;
        }
        listNode.next = l1 == null ? l2 : l1;

        return head;
    }

    /**
     * 解法二：递归法
     */
    public ListNode mergeTwoList2(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        } else if (l2 == null) {
            return l1;
        } else {
            if (l1.val < l2.val) {
                l1.next = mergeTwoList2(l1.next, l2);
                return l1;
            } else {
                l2.next = mergeTwoList2(l1, l2.next);
                return l2;
            }
        }
    }

Leetcode 24. 两两交换链表中的节点

    // 解法一
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode first = head;
        ListNode second = first.next;
        ListNode temp = second.next;
        head = second;
        while (temp != null && temp.next != null) {
            first.next = temp.next;
            second.next = first;
            first = temp;
            second = first.next;
            temp = second.next;
        }
        first.next = second.next;
        second.next = first;
        return head;
    }
    // 解法二
    public ListNode swapPairs2(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode newHead = head.next;
        head.next = swapPairs2(newHead.next);
        newHead.next = head;
        return newHead;
    }