【Lintcode】098.Sort List

题目：

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

Example

Given 1->3->2->null, sort it to 1->2->3->null.

题解：

　　O(n log n) : 快速排序，归并排序，堆排序

Solution 1 ()

class Solution {
public:
    ListNode *sortList(ListNode *head) {
        if (!head || !head->next) {
            return head;
        }
        
        ListNode* slow = head;
        ListNode* fast = head->next;
        
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode* right = sortList(slow->next);
        slow->next = nullptr;
        ListNode* left = sortList(head);
        
        return merge(left, right);
    }
    
    ListNode* merge(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(-1);
        ListNode* cur = dummy;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        
        if (l1) cur->next = l1;
        if (l2) cur->next = l2;
        return dummy->next;
    }
};

Solutin 2 ()

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode *slow = head, *fast = head, *pre = head;
        while (fast && fast->next) {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        return merge(sortList(head), sortList(slow));
    }
    ListNode* merge(ListNode* l1, ListNode* l2) {
        if (!l1) return l2;
        if (!l2) return l1;
        if (l1->val < l2->val) {
            l1->next = merge(l1->next, l2);
            return l1;
        } else {
            l2->next = merge(l1, l2->next);
            return l2;
        }                                         
    }
};