LeetCode Sort List

题目：

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

代码思想参考http://blog.csdn.net/xudli/article/details/16819545

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; *//** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *sortList(ListNode *head) {        return mergeSort(head);    }private:ListNode *mergeSort(ListNode *head) {if(head == NULL || head->next == NULL)return head; ListNode *slow = head, *fast = head;ListNode *pre_slow = head;//分治法：使用快慢指针进行分组，slow之前为一组，之后为一组；while(fast != NULL && fast->next != NULL) {fast = fast->next->next;pre_slow = slow;slow = slow->next;}pre_slow->next = NULL;ListNode *h1 = mergeSort(head);ListNode *h2 = mergeSort(slow);return merge(h1, h2);}ListNode *merge(ListNode *h1, ListNode *h2) {ListNode *dummy = new ListNode(-1);ListNode *cur = dummy;if(h1 == NULL && h2 == NULL)return NULL;while(h1 != NULL && h2 != NULL) {if(h1->val <= h2->val) {cur->next = h1;h1 = h1->next;}else {cur->next = h2;h2 = h2->next;}cur = cur->next;}//链表h1还有剩余 while(h1 != NULL) {cur->next = h1;h1 = h1->next;cur = cur->next;}//链表h2有剩余 while(h2 != NULL) {cur->next = h2;h2 = h2->next;cur = cur->next;} cur = dummy->next;delete dummy;return cur; }};