【面试准备】letcode—sort list

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

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* merged_list(ListNode* l_head,ListNode* r_head){ListNode* merged_head = NULL;ListNode* merged_tail = NULL;if(l_head == NULL){return r_head;}if(r_head == NULL){return l_head;}while(l_head != NULL && r_head != NULL){if(l_head->val <= r_head->val){if(merged_head == NULL){merged_head = l_head;merged_tail = l_head;}else{merged_tail->next = l_head;merged_tail = merged_tail->next;}l_head = l_head->next;}else{if(merged_head == NULL){merged_head = r_head;merged_tail = r_head;}else{merged_tail->next = r_head;merged_tail = merged_tail->next;}r_head = r_head->next;}}if(l_head != NULL){merged_tail->next = l_head;}else if(r_head != NULL){merged_tail->next = r_head;}return merged_head;}ListNode* sortList(ListNode* head){if(head == NULL || head->next == NULL){return head;}ListNode* node = head ;ListNode* slow = head ;ListNode* fast = head ; while(fast->next != NULL && fast->next->next != NULL){slow = slow->next;fast = fast->next->next;}fast = slow->next;slow->next = NULL;slow = head;node = merged_list(sortList(slow),sortList(fast));return node;}};