LeetCode 148:Sort List

Sort a linked list in O(n log n) time using constant space complexity.

//题目要求：对链表进行排序//解题思路：归并排序，再Merge//归并排序的基本思想是：找到链表的中间节点，然后递归对前半部分和后半部分分别进行归并排序，最后对两个排好序的链表进行Merge class Solution { public: ListNode* sortList(ListNode* head) { if (head == NULL || head->next == NULL)  return head; //快慢指针找到中间节点 ListNode *fast = head, *slow = head; while (fast->next != NULL && fast->next->next != NULL) { fast = fast->next->next; slow = slow->next; } //断开 fast = slow; slow = slow->next; fast->next = NULL; ListNode *l1 = sortList(head); //前半段排序 ListNode *l2 = sortList(slow); //后半段排序 return mergeTwoLists(l1, l2); } //Merge Two Sorted Lists ListNode *mergeTwoLists(ListNode *l1, ListNode *l2){ ListNode dummy(-1); for (ListNode* p = &dummy; l1 != NULL || l2 != NULL; p = p->next) { int val1 = l1 == NULL ? INT_MAX : l1->val; int val2 = l2 == NULL ? INT_MAX : l2->val; if (val1 <= val2) { p->next = l1; l1 = l1->next; } else { p->next = l2; l2 = l2->next; } } return dummy.next; } };