问题描述:
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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
分析:
初步想法:
构建一个新的链表,遍历比较L1和L2上的节点,小的加入新的链,然后指向下一个节点,直到有一条链的节点为空,把剩余的链的部分接到新的链上。
代码如下:
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1 == null) return l2;
if(l2 == null) return l1;
ListNode first = new ListNode(0);
ListNode res = first;
while(l1 != null && l2 != null){
if(l1.val <= l2.val) {
res.next = new ListNode(l1.val);
l1 = l1.next;
}else{
res.next = new ListNode(l2.val);
l2 = l2.next;
}
res = res.next;
}
if(l1 == null) res.next = l2;
else res.next = l1;
return first.next;
}
方法二:(递归)
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1 == null) return l2;
if(l2 == null) return l1;
if(l1.val < l2.val){
l1.next = mergeTwoLists(l1.next, l2);
return l1;
}else{
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
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