[LintCode] Merge Two Sorted Lists

Merge Two Sorted Lists

Merge two sorted (ascending) linked lists and return it as a new sorted list. The new sorted list should be made by splicing together the nodes of the two lists and sorted in ascending order.

Example

Given

1->3->8->11->15->null

,

2->null

, return

1->2->3->8->11->15->null

.

SOLUTION:

链表在merge的时候其实要有优势的，它不需要额外空间，不想array，要重新开一个空间放array。操作的时候也很简单，弄一个dummy，然后把小的直接连上就行了。此题属于基本操作，在其他题里完全可以作为function调用。

代码：

/**
* Definition for ListNode.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int val) {
*         this.val = val;
*         this.next = null;
*     }
* }
*/
public class Solution {
/**
* @param ListNode l1 is the head of the linked list
* @param ListNode l2 is the head of the linked list
* @return: ListNode head of linked list
*/
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null){
return l2;
}
if (l2 == null){
return l1;
}
ListNode dummy = new ListNode(0);
ListNode node = dummy;
while (l1 != null && l2 != null){
if (l1.val < l2.val){
node.next = new ListNode(l1.val);
l1 = l1.next;
node = node.next;
} else {
node.next = new ListNode(l2.val);
l2 = l2.next;
node = node.next;
}
}
while (l1 != null){
node.next = new ListNode(l1.val);
l1 = l1.next;
node = node.next;
}
while (l2 != null){
node.next = new ListNode(l2.val);
l2 = l2.next;
node = node.next;
}
return dummy.next;
}
}

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