21.Merge Two Sorted Lists 、23. Merge k Sorted Lists

21.Merge Two Sorted Lists

初始化一个指针作为开头，然后返回这个指针的next

class Solution {
public:ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {ListNode* dummy = new ListNode(-1);ListNode* p = dummy;while(l1 && l2){if(l1->val <= l2->val){p->next = l1;p = p->next;l1 = l1->next;}else{p->next = l2;p = p->next;l2 = l2->next;}}if(l1)p->next = l1;elsep->next = l2;return dummy->next;}
};

 

 

 

23. Merge k Sorted Lists

 

https://www.cnblogs.com/grandyang/p/4606710.html

这个是分治的思想

实质上就是每次合并一半的链表，且两两合并的链表按照一定间隔距离进行合并

class Solution {
public:ListNode* mergeKLists(vector<ListNode*>& lists) {int n = lists.size();if(n <= 0)return NULL;while(n > 1){int k = (n + 1)/2;for(int i = 0;i < n/2;i++)lists[i] = mergeList(lists[i],lists[i + k]);n = k;}return lists[0];}ListNode* mergeList(ListNode* l1,ListNode* l2){if(l1 == NULL)return l2;if(l2 == NULL)return l1;ListNode* head;if(l1->val < l2->val){head = l1;head->next = mergeList(l1->next,l2);}else{head = l2;head->next = mergeList(l1,l2->next);}return head; }
};

自己写的：

用非递归也可以合并两个链表。

k = (n + 1)/2中k代表间隔，vector中的链表等间隔合并，这样能达到减少一半的目的。+1的目的是针对奇数这种情况，中间一定会剩下一个单独的，这个单独的也要保留在vector中。n代表当前已更新剩下的链表个数，其实也就是存放在lists中的前n个。+1的目的其实也是针对奇数个的情况。

class Solution {
public:ListNode* mergeKLists(vector<ListNode*>& lists) {if(lists.empty())return NULL;int n = lists.size();while(n > 1){int k = (n + 1)/2;for(int i = 0;i < n/2;i++)lists[i] = merge(lists[i],lists[i + k]);n = (n + 1)/2;}return lists[0];}ListNode* merge(ListNode* l1,ListNode* l2){ListNode* dummy = new ListNode(-1);ListNode* p = dummy;while(l1 && l2){if(l1->val < l2->val){p->next = l1;p = p->next;l1 = l1->next;}else{p->next = l2;p = p->next;l2 = l2->next;}}if(l1)p->next = l1;elsep->next = l2;return dummy->next;}
};