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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
分析:改题目应该叫合并两个有序的链表,感觉这样才更准确一点。我们的初始想法无非就是再另新建一个链表,然后比大小,存入即可。
首先定义链表:
public class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
}
}
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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1); //这个用-1或者其他值初始化没有关系的。
ListNode cur = dummy;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
cur.next = l1;
l1 = l1.next;
} else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
cur.next = (l1 != null) ? l1 : l2;
return dummy.next; //之所以是返回dummy.next是因为我们最初创建了一个节点,值为-1,所以呢这里要返回dummy.next.
}
还有使用递归的方法如下:
public ListNode mergeTwoLists2(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists2(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists2(l1, l2.next);
return l2;
}
}
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