leetcode 148 —— Sort List

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

思路：归并排序

class Solution {public:ListNode* sortList(ListNode* head) {if (head == nullptr || head->next == nullptr){return head;}ListNode* mid = MoveToMid(head);head = MergeSort(head, mid);return head;}ListNode* MoveToMid(ListNode* head){ListNode *p1 = head, *p2 = head;while (p1 != nullptr && (p2->next&&p2->next->next)){p1 = p1->next;p2 = p2->next->next;}if (p1 != nullptr){ListNode*tmp = p1->next;p1->next = nullptr;//printList(tmp);return tmp;}elsereturn nullptr;}ListNode* MergeSort(ListNode* list1, ListNode* list2){if (list1 == nullptr&&list2 == nullptr)return nullptr;else if ((list1 == nullptr&&list2 != nullptr) ||(list1 != nullptr&&list2 == nullptr)){return list1 == nullptr ? list2 : list1;}ListNode* mid1 = MoveToMid(list1);ListNode* mid2 = MoveToMid(list2);list1 = MergeSort(list1, mid1);list2 = MergeSort(list2, mid2);ListNode* pre = new ListNode(0);ListNode* p = pre;while (list1 != nullptr&&list2 != nullptr){if (list1->val > list2->val){p->next = list2;p = p->next;list2 = list2->next;}else{p->next = list1;p = p->next;list1 = list1->next;}}if (list1 != nullptr)p->next = list1;else if (list2 != nullptr)p->next = list2;else p->next = nullptr;//printList(pre->next);return pre->next;delete pre;}}a;