Merge K Sorted Lists (Leetcode 2

外排序，及k路归并算法是一个很重要的算法，大数据用也会用到。
http://shmilyaw-hotmail-com.iteye.com/blog/1776132

此题，准备用两种方法来解。第一种是用Priority Queue, 简单好想。注意用了c++ Lamda Function, 来定义pque的compare function。

ListNode *mergeKLists(vector<ListNode *> &lists) {
        // write your code here
        if(lists.empty()) return NULL;
        
        auto comp = [](const ListNode *l1, const ListNode *l2){
            return l1->val > l2->val;    
        };
        priority_queue<ListNode*, vector<ListNode*>, decltype(comp)> pque(comp);
        
        ListNode *dummy = new ListNode(0);
        ListNode *pre = dummy;
        
        for(ListNode* it : lists){
            if(it != NULL) pque.push(it);
        }
        while(!pque.empty()){
            ListNode *cur = pque.top(); pque.pop();
            pre->next = cur;
            pre = pre->next;
            if(cur->next) pque.push(cur->next);
        }
        pre = dummy->next;
        delete dummy;
        return pre;
    }

第二种是Divide and Conquer，这种难想一些, 但实现起来并不复杂。
我们把k个list从中间分一半，先合并上k/2 (0 - k/2), 再合并下k/2 (k/2+1 - k)，最后再merge。对上一半，下一半也分别采用相同的方法，先切一半，再合起来。MergeSort的思想。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(lists.empty()) return NULL;
        int size = lists.size();
        return sortList(lists, 0, size-1);
    }
    
    ListNode *sortList(vector<ListNode*>& lists, int start, int end){
        if(start > end) return NULL;
        else if(start == end){
            return lists[start];
        }
        int mid = start + (end-start)/2;
        return merge(sortList(lists, start, mid), sortList(lists, mid+1, end));
    }
    
    ListNode *merge(ListNode *headA, ListNode *headB){
        if(!headA) return headB;
        else if(!headB) return headA;
        ListNode *dummy = new ListNode(0);
        ListNode *pre = dummy;
        while(headA && headB){
            if(headA->val < headB->val){
                pre->next = headA;
                headA = headA->next;
            }
            else{
                pre->next = headB;
                headB = headB->next;
            }
            pre = pre->next;
        }
        pre->next = headA ? headA : headB;
        pre = dummy->next;
        delete dummy;
        return pre;
    }
};