题目源自于leetcode。
题目:Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
解法一:每次在各个有序链表中寻找的表头寻找出最小的一个结点,将其链到新链表上。空间复杂度为O(1),时间复杂度为O(k*n),k为有序链表的个数,n为每个链表的长度。
/**
* 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 n = lists.size();
ListNode *result, *p;
int j;
int min_idx;
int min_value = INT_MAX;
for(j=0;j<n;j++)
{
if(lists[j] != NULL && lists[j]->val < min_value)
{
min_value = lists[j]->val;
min_idx = j;
}
}
if(min_value == INT_MAX)
return NULL;
result = lists[min_idx];
lists[min_idx] = lists[min_idx]->next;
p = result;
p->next = NULL;
while(true)
{
min_value = INT_MAX;
for(j=0;j<n;j++)
{
if(lists[j] != NULL && lists[j]->val < min_value)
{
min_value = lists[j]->val;
min_idx = j;
}
}
if(min_value == INT_MAX)
break;
p->next = lists[min_idx];
lists[min_idx] = lists[min_idx]->next;
p = p->next;
p->next = NULL;
}
return result;
}
};
解法二:每次都是两两合并,合并之后再拿新的过来合并。其中两两合并的函数和两个有序链表的合并 Merge Two Sorted Lists中是一样的。方法二的时空复杂度和方法一是一致的。
class Solution {
public:
ListNode *mergeKLists(vector<ListNode *> &lists) {
if(lists.empty())
return NULL;
int n = lists.size();
ListNode *p = lists[0];
for(int i=1;i<n;i++)
p = mergeTwoLists(p, lists[i]);
return p;
}
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
if(l1 == NULL)
return l2;
if(l2 == NULL)
return l1;
ListNode tmp(0);
ListNode *p = &tmp;
while(l1 != NULL && l2 != NULL)
{
if(l1->val < l2->val)
{
p->next = l1;
l1 = l1->next;
}
else
{
p->next = l2;
l2 = l2->next;
}
p = p->next;
}
if(l1 == NULL)
p->next = l2;
else
p->next = l1;
return tmp.next;
}
};