current position:Home>【LeetCode - Java】21. Merge two ordered linked lists (simple)
【LeetCode - Java】21. Merge two ordered linked lists (simple)
2022-01-26 22:10:04 【Beeemo】
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1. Title Description
2. Their thinking
This kind of topic belongs to classic data structure Class title , understand Linked list After the feature, it's good to start . First Conventional thinking There are two kinds of , One is Make elements in a list Insert To another linked list to form an ordered linked list that meets the requirements ; another The train of thought is Generate a new linked list , Put the elements that meet the requirements in the two old linked lists add to To the new linked list .
Plug in processing relative Generative expressions are a little more troublesome , You need to determine Which old linked list to use as the benchmark , According to the requirements of this topic, you should choose Small opening element The linked list is used as the insertion benchmark , Before that, we need to Judge not empty , Otherwise it will appear Null pointer exception .
in addition to , This topic can also be based on recursive The mind of . Recursive function Is to find out At present Nodal Elements as well as The subsequent Of Sublist , This kind of thinking is actually Be similar to Plug in processing ideas .
3. Code implementation
3.1 Plug in
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode pre;
ListNode list;
if(list1==null)
return list2;
else if(list2==null)
return list1;
else if (list1.val < list2.val) {
list = list1;
list1 = list1.next;
} else {
list = list2;
list2 = list2.next;
}
pre = list;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
pre.next = list1;
list1 = list1.next;
} else {
pre.next = list2;
list2 = list2.next;
}
pre = pre.next;
}
pre.next = list1 == null ? list2 : list1;
return list;
}
3.2 Generative
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode pre;
ListNode list = new ListNode();
pre = list;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
pre.next = list1;
list1 = list1.next;
} else {
pre.next = list2;
list2 = list2.next;
}
pre = pre.next;
}
pre.next = list1 == null ? list2 : list1;
return list.next;
}
3.3 Recursive
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
if (list1 == null)
return list2;
else if (list2 == null)
return list1;
else {
if (list1.val < list2.val) {
list1.next = mergeTwoLists(list1.next, list2);
return list1;
} else {
list2.next = mergeTwoLists(list1, list2.next);
return list2;
}
}
}
This implementation is the subject Relatively optimal performance Solution method .
3.4 contrast
Of the above three methods Time complexity All are O(m+n),m and n They are the of two linked lists length , Therefore, the time performance of the three algorithms There is no difference . And in the Spatial complexity On , Generative algorithm because it needs to generate New linked list , Therefore, compared with other algorithms in terms of space occupation Bigger . Recursive Although the solution is feasible in this subject , Because the number of nodes in this topic does not exceed 50, But if the number of nodes is too large Prone to stack overflow The situation of .
copyright notice
author[Beeemo],Please bring the original link to reprint, thank you.
https://en.cdmana.com/2022/01/202201262210006146.html
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