q21_合并两个有序链表

package 递归.q21_合并两个有序链表.f1;
public class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}
package 递归.q21_合并两个有序链表.f1;
/**
 * 插队法 - 遍历迭代 o(n)
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        ListNode head = new ListNode(Integer.MIN_VALUE);
        head.next = l1;
        ListNode pre = head;
        while (l2 != null) {
            ListNode t1 = pre.next;
            ListNode t2 = l2.next;
            while (l2.val > t1.val) {
                if (t1.next == null) {
                    t1.next = l2;
                    return head.next;
                } else {
                    pre = pre.next;
                    t1 = t1.next;
                }
            }
            pre.next = l2;
            l2.next = t1;
            l2 = t2;
        }
        return head.next;
    }
}

package 递归.q21_合并两个有序链表.f2;
public class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}
package 递归.q21_合并两个有序链表.f2;
/**
 * 递归（看成两个链表头部较小的一个与剩下元素的 merge 操作结果合并） o(n)
 */
public 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;
        }
    }
}