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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.
题目的意思是给定两个有序的链表,将它们合并为一个有序的链表。
思路比较简单,依次比较两个节点值得大小,每次取值小的元素挂到新的链表上,下面分别列出递归实现的代码和迭代实现的代码:
递归实现:
迭代实现:
题目的意思是给定两个有序的链表,将它们合并为一个有序的链表。
思路比较简单,依次比较两个节点值得大小,每次取值小的元素挂到新的链表上,下面分别列出递归实现的代码和迭代实现的代码:
递归实现:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if(l1 == null) return l2; if(l2 == null) return l1; ListNode helper = null; if(l1.val > l2.val) { helper = l2; helper.next = mergeTwoLists(l1, l2.next); } else { helper = l1; helper.next = mergeTwoLists(l1.next, l2); } return helper; } }
迭代实现:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if(l1 == null) return l2; if(l2 == null) return l1; ListNode helper = new ListNode(0); ListNode head1 = l1; ListNode head2 = l2; while(l1 != null && l2 != null) { if(l1.val > l2.val) { helper.next = l2; l2 = l2.next; } else { helper.next = l1; l1 = l1.next; } helper = helper.next; } while (l1 != null) { helper.next = l1; l1 = l1.next; helper = helper.next; } while(l2 != null) { helper.next = l2; l2 = l2.next; helper = helper.next; } return (head1.val > head2.val) ? head2 : head1; } }
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