【Sort List】cpp

题目:

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

代码:

/**
 * 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 || !head->next ) return head;
        ListNode dummy(-1);
        dummy.next = head;
        ListNode *p1=&dummy, *p2=&dummy;
        while ( p2 && p2->next && p2->next->next )
        {
            p1 = p1->next;
            p2 = p2->next->next;
        }
        ListNode *h1 =  Solution::sortList(p1->next);
        p1->next = NULL;
        ListNode *h2 =  Solution::sortList(dummy.next);
        return Solution::mergeTwo(h1, h2);
    }
    static ListNode* mergeTwo(ListNode *h1, ListNode *h2)
    {
        ListNode dummy(-1);
        ListNode *p = &dummy;
        while ( h1 && h2 )
        {
            if ( h1->val<h2->val )
            {
                p->next = h1;
                h1 = h1->next;
            }
            else
            {
                p->next = h2;
                h2 = h2->next;
            }
            p = p->next;
        }
        p->next = h1 ? h1 : h2;
        return dummy.next;
    }
};

tips:

单链表时间要求O(nlongn) 且const extra space,可以选择归并排序(另,双向链表适合用快速排序)

第一次没有AC,原因是少考虑一种返回条件,即“head只有一个元素的时候需要直接返回”,修改之后第二次AC了。

===================================================

第二次过这道题,尝试着摸索写出来,一次AC了。

/**
 * 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 ) return NULL;
            if ( !head->next ) return head;
            ListNode dummpy(0);
            ListNode* p1 = &dummpy;
            ListNode* p2 = &dummpy;
            dummpy.next = head;
            while ( p2 && p2->next )
            {
                p1 = p1->next;
                p2 = p2->next->next;
            }
            ListNode* r = Solution::sortList(p1->next);
            p1->next = NULL;
            ListNode* l = Solution::sortList(dummpy.next);
            return Solution::merge2SortedLists(l,r);

        }
        static ListNode* merge2SortedLists(ListNode* p1, ListNode* p2)
        {
            ListNode head(0);
            ListNode* p = &head;
            while ( p1 && p2 )
            {
                if ( p1->val < p2->val )
                {
                    p->next = p1;
                    p1 = p1->next;
                }
                else
                {
                    p->next = p2;
                    p2 = p2->next;
                }
                p = p->next;
            }
            p->next = p1 ? p1 : p2;
            return head.next;
        }
};

 

posted on 2015-05-18 20:36  承续缘  阅读(284)  评论(0编辑  收藏  举报

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