Leetcode 23. Merge k Sorted Lists

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

题目意思：k个有序链表合并后仍为一个有序链表

解题思路：我先将链表存储在vector中，然后我用新的vector去存储链表中的每个节点，之后快速排序（stl std::sort）,然后再对排序过后的节点组成链表，不是最优解，但是一个比较容易想的方法，时间复杂度是O(kn*logkn)

#include <stdio.h>

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

#include <vector>
#include <algorithm>

bool cmp(const ListNode *a, const ListNode *b){
    return a->val < b->val;
}

class Solution {
public:
    ListNode* mergeKLists(std::vector<ListNode*>& lists) {
        std::vector<ListNode *> node_vec;        
        for (int i = 0; i < lists.size(); i++){
            ListNode *head = lists[i];
            while(head){
                node_vec.push_back(head);
                head = head->next;
            }
        }
        if (node_vec.size() == 0){
            return NULL;
        }        
        std::sort(node_vec.begin(), node_vec.end(), cmp);
        for (int i = 1; i < node_vec.size(); i++){
            node_vec[i-1]->next = node_vec[i];
        }
        node_vec[node_vec.size()-1]->next = NULL;
        return node_vec[0];
    }
};

int main(){
    ListNode a(1);
    ListNode b(4);
    ListNode c(6);
    ListNode d(0);
    ListNode e(5);
    ListNode f(7);
    ListNode g(2);
    ListNode h(3);
    a.next = &b;
    b.next = &c;    
    d.next = &e;
    e.next = &f;    
    g.next = &h;    
    Solution solve;    
    std::vector<ListNode *> lists;
    lists.push_back(&a);
    lists.push_back(&d);
    lists.push_back(&g);    
    ListNode *head = solve.mergeKLists(lists);
    while(head){
        printf("%d
", head->val);
        head = head->next;
    }
    return 0;
}

通过~