0021. Merge Two Sorted Lists (E)
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Merge Two Sorted Lists (E)
题目
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
题意
将两个有序链表合并为一个有序链表。
思路
归并排序。
代码实现
Java
迭代
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0);
ListNode pointer = head;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
pointer.next = l1;
l1 = l1.next;
} else {
pointer.next = l2;
l2 = l2.next;
}
pointer = pointer.next;
}
if (l1 == null) {
pointer.next = l2;
}
if (l2 == null) {
pointer.next = l1;
}
return head.next;
}
}
递归
class Solution {
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;
}
}
}
javascript
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var mergeTwoLists = function (l1, l2) {
let dummy = new ListNode(0)
let 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 ? l1 : l2
return dummy.next
}
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