剑指Offer面试题17 & Leetcode21

剑指Offer面试题17 & Leetcode21

Merge Two Sorted Lists  合并两个排序的链表

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.

解题思路

  考虑：链表1的头结点和链表2的头结点中的较小者为合并后链表的头结点，之后合并的是链表1第二个节点开始的链表和链表2，可以用递归的思想解决问题。注意处理空指针问题。当然也可以用循环，用一个ListNode保留头节点，然后用另一个ListNode遍历两个链表。

Solution1 递归

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {          if(l1 == null)            return l2;        if(l2 == null)            return l1;        ListNode head;        if(l1.val<l2.val){            head = l1;            head.next = mergeTwoLists(l1.next, l2);        }else{            head = l2;            head.next = mergeTwoLists(l1, l2.next);        }        return head;}

Solution2 循环

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {          if (l1 == null) {               return l2;           }           if (l2 == null) {               return l1;           }           ListNode node = null;           ListNode head = null;           while (l1 != null && l2 != null) {               if (l1.val <= l2.val) {                   if (node == null) {                       node = l1;                       head = node;                   } else {                       node.next = l1;                       node = node.next;                   }                   l1 = l1.next;               } else {                   if (node == null) {                       node = l2;                       head = node;                   } else {                       node.next = l2;                       node = node.next;                   }                   l2 = l2.next;          }           }           if (l1 != null) {               node.next = l1;           } else if (l2 != null) {               node.next = l2;           }           return head;      }