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148. Sort List

148. Sort List

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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
// split O(logn)
ListNode* sortList(ListNode* head) {
// head->next 是为了单个节点的情况
if(!head || !head->next)
return head;
auto midPoint = getMid(head);
auto left = sortList(head);
auto right = sortList(midPoint);
return mergeList(left, right);
}
private:
// 这里的获取终点函数记得要进行部分调整,要从中间切断这个list
// O(n)
ListNode* getMid(ListNode* head)
{
ListNode* walker = nullptr;
ListNode* runner = head;
while(runner && runner->next)
{
walker = !walker ? head : walker->next;
runner = runner->next->next;
}
ListNode* mid = walker->next;
walker->next = nullptr;
return mid;
}
// O(n)
ListNode* mergeList(ListNode* l1, ListNode* l2)
{
if(!l1)
return l2;
if(!l2)
return l1;
if(l1->val > l2->val)
{
l2->next = mergeList(l2->next, l1);
return l2;
}else
{
l1->next = mergeList(l1->next, l2);
return l1;
}
return nullptr;
}
};

T(n) : O(nlogn)

总体是分成许多一半一半的,然后每一小部分merge

优化merge
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* sortList(ListNode* head) {
if(!head || !head->next)
return head;
auto midPoint = getMid(head);
auto left = sortList(head);
auto right = sortList(midPoint);
return mergeList(left, right);
}
private:
ListNode* getMid(ListNode* head)
{
ListNode* walker = nullptr;
ListNode* runner = head;
while(runner && runner->next)
{
walker = !walker ? head : walker->next;
runner = runner->next->next;
}
ListNode* mid = walker->next;
walker->next = nullptr;
return mid;
}
ListNode* mergeList(ListNode* l1, ListNode* l2)
{
ListNode dummyHead(0);
auto prev = &dummyHead;
while(l1 && l2)
{
if(l1->val > l2->val)
{
prev->next = l2;
l2 = l2->next;
}else
{
prev->next = l1;
l1 = l1->next;
}
prev = prev->next;
}
if(l1)
prev->next = l1;
else
prev->next = l2;
return dummyHead.next;
}
};