Sort a linked list in O(n log n) time using constant space complexity.
常见排序方法有很多,插入排序,选择排序,堆排序,快速排序,冒泡排序,归并排序,桶排序等等。。它们的时间复杂度不尽相同,而这里题目限定了时间必须为O(nlgn),符合要求只有快速排序,归并排序,堆排序,而根据单链表的特点,最适于用归并排序。代码如下:
C++ 解法一:
class Solution { public: ListNode* sortList(ListNode* head) { if (!head || !head->next) return head; ListNode *slow = head, *fast = head, *pre = head; while (fast && fast->next) { pre = slow; slow = slow->next; fast = fast->next->next; } pre->next = NULL; return merge(sortList(head), sortList(slow)); } ListNode* merge(ListNode* l1, ListNode* l2) { ListNode *dummy = new ListNode(-1); ListNode *cur = dummy; while (l1 && l2) { if (l1->val < l2->val) { cur->next = l1; l1 = l1->next; } else { cur->next = l2; l2 = l2->next; } cur = cur->next; } if (l1) cur->next = l1; if (l2) cur->next = l2; return dummy->next; } };
Java 解法一:
public class Solution { public ListNode sortList(ListNode head) { if (head == null || head.next == null) return head; ListNode slow = head, fast = head, pre = head; while (fast != null && fast.next != null) { pre = slow; slow = slow.next; fast = fast.next.next; } pre.next = null; return merge(sortList(head), sortList(slow)); } public ListNode merge(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(-1); ListNode cur = dummy; while (l1 != null && l2 != null) { if (l1.val < l2.val) { cur.next = l1; l1 = l1.next; } else { cur.next = l2; l2 = l2.next; } cur = cur.next; } if (l1 != null) cur.next = l1; if (l2 != null) cur.next = l2; return dummy.next; } }
下面这种方法也是归并排序,而且在merge函数中也使用了递归,这样使代码更加简洁啦~
C++ 解法二:
class Solution { public: ListNode* sortList(ListNode* head) { if (!head || !head->next) return head; ListNode *slow = head, *fast = head, *pre = head; while (fast && fast->next) { pre = slow; slow = slow->next; fast = fast->next->next; } pre->next = NULL; return merge(sortList(head), sortList(slow)); } ListNode* merge(ListNode* l1, ListNode* l2) { if (!l1) return l2; if (!l2) return l1; if (l1->val < l2->val) { l1->next = merge(l1->next, l2); return l1; } else { l2->next = merge(l1, l2->next); return l2; } } };
Java 解法二:
public class Solution { public ListNode sortList(ListNode head) { if (head == null || head.next == null) return head; ListNode slow = head, fast = head, pre = head; while (fast != null && fast.next != null) { pre = slow; slow = slow.next; fast = fast.next.next; } pre.next = null; return merge(sortList(head), sortList(slow)); } public ListNode merge(ListNode l1, ListNode l2) { if (l1 == null) return l2; if (l2 == null) return l1; if (l1.val < l2.val) { l1.next = merge(l1.next, l2); return l1; } else { l2.next = merge(l1, l2.next); return l2; } } }
本文转自博客园Grandyang的博客,原文链接:链表排序[LeetCode] Sort List ,如需转载请自行联系原博主。