LeetCode之“链表”:Merge Two Sorted Lists && Merge k Sorted Lists
1. Merge Two Sorted Lists
题目要求:
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.
这道题目题意是要将两个有序的链表合并为一个有序链表。为了编程方便,在程序中引入dummy节点。具体程序如下:
/**
* 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 *dummy = new ListNode();
ListNode *head = nullptr, *start = dummy;
int flag = true;
while(l1 && l2)
{
if(l1->val < l2->val)
{
start->next = l1;
l1 = l1->next;
}
else
{
start->next = l2;
l2 = l2->next;
}
start = start->next;
} if(!l1)
start->next = l2;
else if(!l2)
start->next = l1; head = dummy->next;
delete dummy;
dummy = nullptr; return head;
}
};
2. Merge k Sorted Lists
题目要求:
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
这道题本来想用比较直观的方法,但超时了。具体程序如下:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
int sz = lists.size();
if(sz == )
return nullptr; ListNode *node = new ListNode();
ListNode *head = nullptr, *start = node;
while(true)
{
int count = ;
int minVal = INT_MAX;
int minNode = -;
for(int i = ; i < sz; i++)
{
if(lists[i] && lists[i]->val < minVal)
{
minNode = i;
minVal = lists[i]->val;
}
else if(!lists[i])
count++;
} if(count != sz)
{
start->next = lists[minNode];
lists[minNode] = lists[minNode]->next;
}
else
{
head = node->next;
delete node;
node = nullptr; return head;
} start = start->next;
}
}
};
后来参考网上的做法,用Divide and Conquer方法会比较好,具体程序如下:
/**
* 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();
ListNode *head = nullptr, *start = node;
int flag = true;
while(l1 && l2)
{
if(l1->val < l2->val)
{
start->next = l1;
l1 = l1->next;
}
else
{
start->next = l2;
l2 = l2->next;
}
start = start->next;
} if(!l1)
start->next = l2;
else if(!l2)
start->next = l1; head = node->next;
delete node;
node = nullptr; return head;
} ListNode *mergeKLists(vector<ListNode*>& lists, int low, int high)
{
if(low < high)
{
int mid = (low + high) / ;
return mergeTwoLists(mergeKLists(lists, low, mid), mergeKLists(lists, mid + , high));
}
return lists[low];
} ListNode* mergeKLists(vector<ListNode*>& lists) {
int sz = lists.size();
if(sz == )
return nullptr; return mergeKLists(lists, , sz - );
}
};
算法分析摘自同一博文:
我们来分析一下上述算法的时间复杂度。假设总共有k个list,每个list的最大长度是n,那么运行时间满足递推式T(k) = 2T(k/2)+O(n*k)。根据主定理,可以算出算法的总复杂度是O(nklogk)。如果不了解主定理的朋友,可以参见主定理-维基百科。空间复杂度的话是递归栈的大小O(logk)。
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