Leetcode 21 Merge Two Sorted Lists
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Merge Two Sorted Lists
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Solution1
- 这道题思路特别清晰,就是分别从两条链表的头部开始,每次取最小的,并将最小的往后挪一位。如此往复。代码如下:
public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(-1);//利用虚节点好处多多 ListNode node = dummy; while(l1!=null&&l2!=null){ if(l1.val<l2.val){ node.next = new ListNode(l1.val); l1 = l1.next; node = node.next; } else{ node.next = new ListNode(l2.val); l2 = l2.next; node = node.next; } } node.next = l1==null?l2:l1; return dummy.next; }}
Solution2
- 解法一相当于是重新造了一条新的链表。每次都将较小的节点链接在已有链表的末尾。但是仔细一想其实不用创建出一条新的链表,可以直接在原有节点上进行操作,只是在搭建链表的时候注意暂存好后面的节点即可。这里的主要思路是:第一条链表上插入第二条链表中较小的节点。代码如下:
public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(-1); dummy.next = l1; ListNode node = dummy; for(;node.next!=null&&l2!=null;node=node.next){//这里实际上是每次将第二条链表的较小节点插入到第一条链表中 if(node.next.val>l2.val){ ListNode temp = node.next;//暂存好后面的节点 node.next = l2;//将链表搭向较小的节点上 l2 = l2.next;//第二条链表往后挪一位 node.next.next = temp;//重新将第一条链表后面的节点搭建回来 } } if(l2!=null) node.next = l2;//若第二条链表还有剩余直接搭建在后面即可 return dummy.next; }}
Solution3
- 除了上面两种迭代的解法外,还可以用递归的方式去求解。代码更加的简洁易懂。
public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if(l1==null||l2==null) return l1==null?l2:l1; if(l1.val<l2.val){ l1.next = mergeTwoLists(l1.next,l2); return l1; }else{ l2.next = mergeTwoLists(l1,l2.next); return l2; } }}
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