问题描述
Sort a linked list in O(n log n) time using constant space complexity.
算法分析
1、要求时间复杂度为 O(n log n),可以考虑归并与快排;
2、本文使用归并,每次将链表从中间位置切断,一分为二;
3、递归2过程,直到链表长度为1;
C++实现
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
ListNode* mergeList(ListNode* head1, ListNode* head2) {
shared_ptr<ListNode> newHead(new ListNode(0));
ListNode* p = newHead.get();
while (head1 && head2) {
if (head1->val <= head2->val) {
p->next = head1;
head1 = head1->next;
}
else {
p->next = head2;
head2 = head2->next;
}
p = p->next;
}
p->next = !head1 ? head2 : head1;
return newHead->next;
}
public:
ListNode* sortList(ListNode* head) {
if (!head || !head->next)
return head;
//devide into two lists
ListNode* slow = head,
*fast = head->next;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
fast = slow->next;
slow->next = nullptr;
return mergeList(sortList(head), sortList(fast));
}
};
时间复杂度为O(n*log(n)),空间复杂度为O(1)。