lintcode-98-链表排序

98-链表排序

在 O(n log n) 时间复杂度和常数级的空间复杂度下给链表排序。

样例

给出 1->3->2->null，给它排序变成 1->2->3->null.

挑战

分别用归并排序和快速排序做一遍。

标签

链表

思路

采用归并排序（时间复杂度是O(nlogn)的排序有快速排序、归并排序、堆排序），使用快慢指针找出链表中点。

code

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
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) {
        // write your code here
        if(head == NULL || head->next == NULL) {
            return head;
        }

        ListNode *fast = head, *slow = head, *temp = head;;
        while(fast != NULL && fast->next != NULL) {
            temp = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        temp->next = NULL;
        return mergeList(sortList(head), sortList(slow));
    }

    ListNode *mergeList(ListNode *head1, ListNode *head2) {
        if(head1 == NULL) {
            return head2;
        }
        if(head2 == NULL) {
            return head1;
        }

        ListNode newHead(0);
        ListNode *temp = &newHead;
        while(head1 != NULL && head2 != NULL) {
            if(head1->val < head2->val) {
                temp->next = head1;
                head1 = head1->next;
            }
            else {
                temp->next = head2;
                head2 = head2->next;
            }
            temp = temp->next;
        }
        if(head1 != NULL) {
            temp->next = head1;
        }
        else if(head2 != NULL) {
            temp->next = head2;
        }

        return newHead.next;
    }
};