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题目描述
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
对比两个表头的元素,将较小者插入新的链表中。
源代码
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
对比两个表头的元素,将较小者插入新的链表中。
源代码
package leetcode; /*private class ListNode { int val; ListNode next; ListNode(int x) { val = x; next = null; } }*/ public class MergeTwoSortedLists { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { //处理空串的情形 if(l1 == null) return l2; if(l2 == null) return l1; ListNode result = null; ListNode lastNode = null; if(l1.val <= l2.val){ lastNode = new ListNode(l1.val); l1 = l1.next; } else{ lastNode = new ListNode(l2.val); l2 = l2.next; } result = lastNode; while(l1 != null && l2 != null){ ListNode newNode = null; if(l1.val <= l2.val){ newNode = new ListNode(l1.val); l1 = l1.next; } else{ newNode = new ListNode(l2.val); l2 = l2.next; } lastNode.next = newNode; lastNode = newNode; } if(l1 == null) lastNode.next = l2; if(l2 == null) lastNode.next = l1; return result; } public static void main(String[] args) { ListNode node1 = new ListNode(1); ListNode node2 = new ListNode(2); ListNode node3 = new ListNode(3); ListNode node4 = new ListNode(4); ListNode node5 = new ListNode(5); ListNode node6 = new ListNode(6); ListNode node7 = new ListNode(1); ListNode node8 = new ListNode(2); ListNode node9 = new ListNode(3); ListNode node10 = new ListNode(4); ListNode node11 = new ListNode(5); ListNode node12 = new ListNode(6); node1.next = node2; node2.next = node3; node3.next = node4; node4.next = node5; node5.next = node6; node7.next = node8; node8.next = node9; node9.next = node10; node10.next = node11; node11.next = node12; ListNode l1 = node1;//System.out.println(l1.val); ListNode l2 = node7;//System.out.println(l2.val); MergeTwoSortedLists mtsl = new MergeTwoSortedLists(); ListNode result = mtsl.mergeTwoLists(l1, l2); while(result != null){ System.out.print(result.val + " "); result = result.next; } } }
发表评论
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