leetcode第22题--Merge k Sorted Lists

problem：Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

先合并两个list，再根据归并排序的方法递归合并。假设总共有k个list，每个list的最大长度是n，那么运行时间满足递推式T(k) = 2T(k/2)+O(n*k)。根据主定理，可以算出算法的总复杂度是O(nklogk)。空间复杂度的话是递归栈的大小O(logk)。

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

 *     ListNode(int x) : val(x), next(NULL) {}

 * };

 */

class Solution {

public:

ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) // 将两个list  merge

{

    if(!l1)

        return l2;

    if(!l2)

        return l1;

    ListNode *tmp = new ListNode(0);

    ListNode *ans = tmp;

    while(l1 && l2)

    {

        if(l1 -> val < l2 -> val)

        {

            ans -> next = l1;

            ans = ans -> next;

            l1 = l1 -> next;

        }

        else

        {

            ans -> next = l2;

            ans = ans -> next;

            l2 = l2 -> next;

        }

    }

    if (l1)

    {

        while(l1)

        {

            ans -> next = l1;

            ans = ans -> next;

            l1 = l1 -> next;

        }

    }



    if (l2)

    {

        while(l2)

        {

            ans -> next = l2;

            ans = ans -> next;

            l2 = l2 -> next;

        }

    }

    ans = tmp;

    delete ans;

    return tmp -> next;

}

ListNode *reMerge(vector<ListNode *> &lists, int l, int r)

{

    if(l < r) // 类似于归并排序，把一大块分成l和r两块，然后将两个合并（l和r也还可以再分）

    {

        int m = (l + r)/2;

        return mergeTwoLists(reMerge(lists, l, m), reMerge(lists, m + 1, r));

    }

    else

        return lists[l];

}

ListNode *mergeKLists(vector<ListNode *> &lists)

{

    ListNode *ans = NULL;// 用NULL 而不是利用 new ListNode(0)

    if (lists.size() == 0)

        return ans;

    return reMerge(lists, 0, lists.size() - 1);

}

};