[leetcode]148. Sort List

题目链接：https://leetcode.com/problems/sort-list/#/description

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

/** * Merge sort use bottom-up policy,  * so Space Complexity is O(1) * Time Complexity is O(NlgN) * stable sort*/class Solution {public:ListNode *sortList(ListNode *head) {if(!head || !(head->next)) return head;//get the linked list's lengthListNode* cur = head;int length = 0;while(cur){length++;cur = cur->next;}ListNode dummy(0);dummy.next = head;ListNode *left, *right, *tail;for(int step = 1; step < length; step <<= 1){cur = dummy.next;tail = &dummy;while(cur){left = cur;right = split(left, step);cur = split(right,step);tail = merge(left, right, tail);}}return dummy.next;}private:/** * Divide the linked list into two lists,     * while the first list contains first n ndoes * return the second list's head */ListNode* split(ListNode *head, int n){//if(!head) return NULL;for(int i = 1; head && i < n; i++) head = head->next;if(!head) return NULL;ListNode *second = head->next;head->next = NULL;return second;}/**  * merge the two sorted linked list l1 and l2,  * then append the merged sorted linked list to the node head  * return the tail of the merged sorted linked list */ListNode* merge(ListNode* l1, ListNode* l2, ListNode* head){ListNode *cur = head;while(l1 && l2){if(l1->val > l2->val){cur->next = l2;cur = l2;l2 = l2->next;}else{cur->next = l1;cur = l1;l1 = l1->next;}}cur->next = (l1 ? l1 : l2);while(cur->next) cur = cur->next;return cur;}};