Leetcode - Merge Two Sorted Lists
2015-01-14 09:28
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null)
return l2;
else if (l2 == null)
return l1;
else if (l1 == null || l2 == null)
return null;
ListNode ln = new ListNode(0);
ListNode temp = ln;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
temp.next = l1;
l1 = l1.next;
} else {
temp.next = l2;
l2 = l2.next;
}
temp = temp.next;
}
if (l2 != null)
temp.next = l2;
else if (l1 != null)
temp.next = l1;
return ln.next;
}
}
================
链表距离是3, 第一个新Node.NEXT ->LIST ; 第二 List 后移; 第三Node = node.next;
===============
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.
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null)
return l2;
else if (l2 == null)
return l1;
else if (l1 == null || l2 == null)
return null;
ListNode ln = new ListNode(0);
ListNode temp = ln;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
temp.next = l1;
l1 = l1.next;
} else {
temp.next = l2;
l2 = l2.next;
}
temp = temp.next;
}
if (l2 != null)
temp.next = l2;
else if (l1 != null)
temp.next = l1;
return ln.next;
}
}
================
链表距离是3, 第一个新Node.NEXT ->LIST ; 第二 List 后移; 第三Node = node.next;
===============
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.
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