[LintCode] Sort List 链表排序

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

Have you met this question in a real interview?

Yes

Example

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

Challenge

Solve it by merge sort & quick sort separately.

LeetCode上的原题，请参见我之前的博客Sort List。

解法一：

class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: You should return the head of the sorted linked list,
     using constant space complexity.
     */
    ListNode *sortList(ListNode *head) {
        if (!head || !head->next) return head;
        ListNode *fast = head, *slow = 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;
        }
    }
};

解法二：

class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: You should return the head of the sorted linked list,
                    using constant space complexity.
     */
    ListNode *sortList(ListNode *head) {
        if (!head || !head->next) return head;
        ListNode *fast = head, *slow = 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) {
        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;
    }
};