Problem:
Sort a linked list in O(n log n) time using constant space complexity.
解题思路:
首先,时间复杂度能达到O(nlgn)的排序算法,常见的有3种:堆排序、归并排序和高速排序,
而对于链表,用堆排序显然不太可能,所以,我们可用归并或者是快排.因为合并两个链表,仅仅用
改动对应的指针,所以其能做到空间复杂度O(1).以下是利用归并排序思想实现的链表排序.
解题思路:
class Solution { public: ListNode *sortList(ListNode *p) { if (p == NULL || p->next == NULL) return p; //增加一个头节点,避免合并时讨论rear为空的情况. ListNode *head = new ListNode(-1), *q = head; head->next = p; int cnt = 0; while (p) { ++cnt; p = p->next; if (cnt % 2 == 0) q = q->next; } p = q->next, q->next = NULL; //递归进行左右两支排序 head->next = q = sortList(head->next); p = sortList(p); //合并 q = Merge(head, p); free(head); return q; } ListNode* Merge(ListNode *head, ListNode *r) { ListNode *l = head->next, *rear = head; while (l && r) { if (l->val < r->val) { rear->next = l; l = l->next, rear = rear->next; } else { rear->next = r; r = r->next, rear = rear->next; } } while (l) { rear->next = l; l = l->next, rear = rear->next; } while (r) { rear->next = r; r = r->next, rear = rear->next; } return head->next; } };