题:
```` https://leetcode.com/problems/merge-two-sorted-lists/ 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.
解法一:创建一个新的链表:
```java
/**
* 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 && l2 == null) {
return null;
} else if (l1 == null) {
return l2;
} else if (l2 == null) {
return l1;
} else {
ListNode head = null;
ListNode next = new ListNode(0); // before head
while (l1 != null || l2 != null) {
if (l1 != null && l2 != null) {
if (l1.val < l2.val) {
next.next = l1;
l1 = l1.next;
} else {
next.next = l2;
l2 = l2.next;
}
} else if (l1 == null) {
next.next = l2;
l2 = l2.next;
} else {
next.next = l1;
l1 = l1.next;
}
head = head == null ? next.next : head;
next = next.next;
}
return head;
}
}
}
解法二:将l2加入到l1中
/**
* 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 && l2 == null) {
return null;
} else if (l1 == null) {
return l2;
} else if (l2 == null) {
return l1;
} else {
// merge l2 to l1
ListNode head = null;
ListNode prev = null;
while (l2 != null) {
if (l1 == null) {
prev.next = l2;
break;
}
if (l1.val > l2.val) {
if (prev == null) {
prev = l2;
} else {
prev.next = l2;
}
ListNode node = l2.next;
l2.next = l1;
prev = l2;
l2 = node;
} else {
prev = l1;
l1 = l1.next;
}
head = head == null ? prev : head;
}
return head;
}
}
}
作者:ywheel
本文出处:http://blog.ywheel.com/post/2015/03/12/leetcode_21/
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