21. 合并两个有序链表

发布于 19 天前  11 次阅读


将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

输入:l1 = [1,2,4], l2 = [1,3,4]
输出:[1,1,2,3,4,4]

输入:l1 = [], l2 = []
输出:[]

输入:l1 = [], l2 = [0]
输出:[0]

递归解法:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        } else if (list2 == null) {
            return list1;
        } else if (list1.val < list2.val) {
            list1.next = mergeTwoLists(list1.next, list2);
            return list1;
        } else {
            list2.next = mergeTwoLists(list1, list2.next);
            return list2;
        }
    }
}

执行用时:0 ms, 在所有 Java 提交中击败了100.00%的用户

内存消耗:41.1 MB, 在所有 Java 提交中击败了38.22%的用户

迭代解法:

创建一个哨兵节点,指向两个链表中较小的那个。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode preHead = new ListNode();
        ListNode pre = preHead;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                pre.next = list1;
                list1 = list1.next;
            } else {
                pre.next = list2;
                list2 = list2.next;
            }
            pre = pre.next;
        }

        pre.next = list1==null?list2:list1;

        return preHead.next;
    }
}

执行用时:0 ms, 在所有 Java 提交中击败了100.00%的用户

内存消耗:41.2 MB, 在所有 Java 提交中击败了26.74%的用户