21. 合并两个有序链表
将两个升序
链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例 1:
输入: l 1 l1 l1 = [1,2,4], l 2 l2 l2 = [1,3,4]
输出:[1,1,2,3,4,4]
示例 2:
输入: l 1 l1 l1 = [], l 2 l2 l2 = []
输出:[]
示例 3:
输入: l 1 l1 l1 = [], l 2 l2 l2 = [0]
输出:[0]
提示:
- 两个链表的节点数目范围是 [0, 50]
- -100 <=
Node.val
<= 100 - l 1 l1 l1 和 l 2 l2 l2 均按 非递减顺序 排列
思路:
法一:常规比较
法二:递归
代码:(Java、C++)
法一:常规比较
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode newList = new ListNode();
ListNode pre = newList;
while(list1 != null || list2 != null){
if(list1 == null){
pre.next = list2;
break;
}
if(list2 == null){
pre.next = list1;
break;
}
if(list1.val <= list2.val){
pre.next = list1;
list1 = list1.next;
pre = pre.next;
pre.next = null;
}else{
pre.next = list2;
list2 = list2.next;
pre = pre.next;
pre.next = null;
}
}
return newList.next;
}
}
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
ListNode* newList = new ListNode();
ListNode* pre = newList;
while(list1 != NULL || list2 != NULL){
if(list1 == NULL){
pre->next = list2;
break;
}
if(list2 == NULL){
pre->next = list1;
break;
}
if(list1->val <= list2->val){
pre->next = list1;
list1 = list1->next;
pre = pre->next;
pre->next = NULL;
}else{
pre->next = list2;
list2 = list2->next;
pre = pre->next;
pre->next = NULL;
}
}
return newList->next;
}
};
法二:递归
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
if(list1 == null) return list2;
if(list2 == null) return list1;
if(list1.val <= list2.val){
list1.next = mergeTwoLists(list1.next, list2);
return list1;
}else{
list2.next = mergeTwoLists(list1, list2.next);
return list2;
}
}
}
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
if(list1 == NULL) return list2;
if(list2 == NULL) return list1;
if(list1->val <= list2->val){
list1->next = mergeTwoLists(list1->next, list2);
return list1;
}else{
list2->next = mergeTwoLists(list1, list2->next);
return list2;
}
}
};
运行结果:
复杂度分析:
- 时间复杂度: O ( m + n ) O(m+n) O(m+n), m m m, n n n,分别为 l 1 l1 l1、 l 2 l2 l2的长度
- 空间复杂度: 法一: O ( 1 ) 法一:O(1) 法一:O(1); 法二: O ( m + n ) 法二:O(m+n) 法二:O(m+n),对于递归调用 自身
mergeTwoLists()
,当它遇到终止条件准备回溯时,已经递归调用了 m + n m+n m+n次,使用了 m + n m+n m+n个栈帧,故最后的空间复杂度为 O ( m + n ) O(m+n) O(m+n)。
题目来源:力扣。
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