Merge Two Sorted Lists
2016-12-26 07:34
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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.
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思想:合并两个有序的链表,链表的基础操作,以其中的一条链表为基准,大了往后插即可。算法时间复杂度是O(m+n),m和n分别是两条链表的长度,空间复杂度是O(1)
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
ListNode *node = new ListNode(0);
ListNode *temp = node;
while(l1 != NULL && l2 !=NULL){
if(l1->valval){
temp->next=l1;
l1=l1->next;
}else{
temp->next = l2;
l2 = l2->next;
}
temp=temp->next;
}
if(l1 != NULL) temp->next=l1;
else temp->next =l2;
return node->next;
}
};
这个题类似的有Merge
Sorted Array,只是后者是对数组进行合并操作,面试中可能会一起问到。扩展题目Merge
k Sorted Lists, 这是一个在分布式系统中比较有用的基本操作,还是需要重视,面试中可以发散出很多问题。
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.
Subscribe to see which companies asked this question
思想:合并两个有序的链表,链表的基础操作,以其中的一条链表为基准,大了往后插即可。算法时间复杂度是O(m+n),m和n分别是两条链表的长度,空间复杂度是O(1)
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
ListNode *node = new ListNode(0);
ListNode *temp = node;
while(l1 != NULL && l2 !=NULL){
if(l1->valval){
temp->next=l1;
l1=l1->next;
}else{
temp->next = l2;
l2 = l2->next;
}
temp=temp->next;
}
if(l1 != NULL) temp->next=l1;
else temp->next =l2;
return node->next;
}
};
这个题类似的有Merge
Sorted Array,只是后者是对数组进行合并操作,面试中可能会一起问到。扩展题目Merge
k Sorted Lists, 这是一个在分布式系统中比较有用的基本操作,还是需要重视,面试中可以发散出很多问题。
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