[LeetCode] Sort List

 

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

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

对链表进行排序

1. 要求算法复杂度为O(nlogn)，只有使用快速排序，归并排序，堆排序。

2. 选择归并排序。

3. 对链表进行递归划分后合并即可。

递归

// Iterator
/**
 * 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) {
        if (head == NULL || head->next == NULL)
            return head;
        ListNode* fast = head;
        ListNode* slow = head;
        ListNode* prev = NULL;
        while (fast && fast->next)
        {
            prev = slow;
            fast = fast->next->next;
            slow = slow->next;
        }
        prev->next = NULL;
        ListNode* l1 = sortList(head);
        ListNode* l2 = sortList(slow);
        return mergeList(l1, l2);
    }
    ListNode* mergeList(ListNode* l1, ListNode* l2)
    {
        if (l1 == NULL)
            return l2;
        if (l2 == NULL)
            return l1;
        if (l1->val < l2->val)
        {
            l1->next = mergeList(l1->next, l2);
            return l1;
        }
        else
        {
            l2->next = mergeList(l1, l2->next);
            return l2;
        }
    }
};

迭代

// Recursion
/**
 * 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) {
        if (head == NULL || head->next == NULL)
            return head;
        ListNode* fast = head;
        ListNode* slow = head;
        ListNode* prev = NULL;
        while (fast && fast->next)
        {
            prev = slow;
            fast = fast->next->next;
            slow = slow->next;
        }
        prev->next = NULL;
        ListNode* l1 = sortList(head);
        ListNode* l2 = sortList(slow);
        return mergeList(l1, l2);
    }
    ListNode* mergeList(ListNode* l1, ListNode* l2)
    {
        ListNode* dummy = new ListNode(0);
        ListNode* curr = dummy;
        while (l1 && l2)
        {
            if (l1->val < l2->val)
            {
                curr->next = l1;
                l1 = l1->next;
            }
            else
            {
                curr->next = l2;
                l2 = l2->next;
            }
            curr = curr->next;
        }
        if (l1)
            curr->next = l1;
        if (l2)
            curr->next = l2;
        return dummy->next;
    }
};