
思路
- 判断有无空链表
- 定义一个新的头结点MergeHead
- 利用递归法(MergeHead = ? , MergeHead.next = ?)
注:只涉及到指针操作,而不设计节点操作不需要new ListNode(),对比题Romove Nth ListNode of List
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public class Solution{ public ListNode mergeTwoSortedLists(ListNode l1 , ListNode l2){ if(l1 == null) return l2; if(l2 == null) return l1; ListNode MergeHead = null; if(l1.val < l2.val){ MergeHead = l1; MergeHead.next = mergeTwoSortedLists(l1.next,l2); }else{ MergeHead = l2; MergeHead.next = mergeTwoSortedLists(l1,l2.next); } return MergeHead; } }
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