23. Merge k Sorted Lists

题目：Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
思路：将K个有序的链表合并成一个有序的链表，可以利用分而治之的方法去做，将lists中的链表分成两部分，先分别合并这两部分的链表，然后再将这两部分的链表进行合并，这样就将问题逐渐转化为两个有序链表的合并，代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* getMergeLists(vector<ListNode*>& lists, int bgn, int end) {        if (bgn > end) return NULL;        else if (bgn == end) return lists[bgn];        else {            int mid = bgn + (end - bgn) / 2;            ListNode* left = getMergeLists(lists, bgn, mid);            ListNode* right = getMergeLists(lists, mid+1, end);            if (!left) return right;            else if (!right) return left;            ListNode *head = NULL, *pre = NULL;            while (left && right) {                if (left->val < right->val) {                    if (!head) {                        head = pre = left;                    }                    else {                        pre->next = left;                        pre = left;                    }                    left = left->next;                }                else {                    if (!head) {                        head = pre = right;                    }                    else {                        pre->next = right;                        pre = right;                    }                    right = right->next;                }            }            if (!left) pre->next = right;            else pre->next = left;            return head;        }    }    ListNode* mergeKLists(vector<ListNode*>& lists) {        return getMergeLists(lists, 0, lists.size()-1);    }};