[LeetCode] 21. 合并两个有序链表

Posted 2021-02-20 怕什么

一开始写的没有注意到在while中判断的时候需要判断 l1 和 l2 同时不能为空，否则会一直在循环里，且由于某一个链表走到最后以后再取值会报错，初始链表应该用new ListNode（0）来初始化

package leetcode;

/**
 * @author doyinana
 * @create 2020-05-01 13:01
 */
public class L21 {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode head1=l1;
        ListNode head2=l2;
        ListNode head=null;
        ListNode cur=head;
        while(l1!=null||l2!=null){
            if(l1.val<=l2.val){
                cur.next=l1;
                l1=l1.next;
                cur=cur.next;
            }else{
                cur.next=l2;
                l2=l2.next;
                cur=cur.next;
            }
        }

        while (l1!=null){
            cur.next=l1;
            cur=cur.next;
            l1=l1.next;
        }

        while (l2!=null){
            cur.next=l2;
            cur=cur.next;
            l2=l2.next;
        }
        return head;
    }
}

View Code

正确的写法如下：

迭代解法

package leetcode;

/**
 * @author doyinana
 * @create 2020-05-01 13:01
 */
public class L21 {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
            ListNode dummyHead = new ListNode(0);
            ListNode tail = dummyHead;
            while (l1 != null && l2 != null) {
                if (l1.val < l2.val) {
                    tail.next = l1;
                    l1 = l1.next;
                } else {
                    tail.next = l2;
                    l2 = l2.next;
                }
                tail = tail.next;
            }

            tail.next = l1 == null? l2: l1;

            return dummyHead.next;
    }
}

 

 

 递归解法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        if (l1.val <= l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }
        l2.next = mergeTwoLists(l1, l2.next);
        return l2;
    }
}