21. Merge Two Sorted Lists
题目
原始地址:https://leetcode.com/problems/merge-two-sorted-lists/#/description
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.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
}
}
描述
归并两个已经排序好的链表。
分析
题目比较简单,解法很直观,依次选择两个链表较小的节点将其归入新链表,并且向后移动,直到结束即可。
给出循环和递归两种解法。
解法1
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0), curr = head;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
curr.next = l1;
l1 = l1.next;
} else {
curr.next = l2;
l2 = l2.next;
}
curr = curr.next;
}
if (l1 != null) {
curr.next = l1;
} else {
curr.next = l2;
}
return head.next;
}
}
解法2
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
} else if (l2 == null) {
return l1;
} else if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}