leetcode23 -- Merge k Sorted Lists

Merge k Sorted Lists

题目：
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

题意：
合并k个已经排序过的链表。

思路：
1.一次合并两个，合并k-1次，但是超时了 Time Limit Exceeded，o(n^2)级别
2.类似归并排序o(nlogn)
3.利用堆性质。其实就是每次比较所有链表的对应的首节点，然后找到最小的，添加到新链表上，删除原链表的最小那个节点，接着让被删除节点的链表的下一个节点进来参与比较。我们使用堆能将时间复杂度降到nlogn级别，和刚刚说的一样，就是把所有链表的首节点建一个堆，然后在堆中删除最小值时间复杂度是log n。删除后在让下一个进堆。

代码：

思路1：(超时)

/**
 * 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){
        if(l1 == NULL && l2 == NULL){
            return NULL;
        }else if(l1 == NULL && l2 != NULL){
            return l2;
        }else if(l1 != NULL && l2 == NULL){
            return l1;
        }
        ListNode *new_head = NULL;
        if(l1->val < l2->val){
            new_head = l1;
            l1 = l1->next;
        }else{
            new_head = l2;
            l2 = l2->next;
        }
        ListNode *p = new_head;
        while(l1 != NULL && l2 != NULL){
            if(l1->val < l2->val){
                p->next = l1;
                p = p->next;
                l1 = l1->next;
            }else{
                p->next = l2;
                p = p->next;
                l2 = l2->next;
            }
        }
        if(l1 != NULL){
            p->next = l1;
        }else if(l2 != NULL){
            p->next = l2;
        }

        return new_head;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(lists.size() == 0){
            return NULL;
        }else if(lists.size() == 1){
            return lists[0];
        }
        ListNode *ret;
        for(int i = lists.size()-1; i > 0; --i){
            ret = mergeTwoLists(lists[i], lists[i-1]);
            lists[i-1] = ret;
        }
        return lists[0];
    }
};

思路2：(通过)

/**
 * 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){
        if(l1 == NULL && l2 == NULL){
            return NULL;
        }else if(l1 == NULL && l2 != NULL){
            return l2;
        }else if(l1 != NULL && l2 == NULL){
            return l1;
        }
        ListNode *new_head = NULL;
        if(l1->val < l2->val){
            new_head = l1;
            l1 = l1->next;
        }else{
            new_head = l2;
            l2 = l2->next;
        }
        ListNode *p = new_head;
        while(l1 != NULL && l2 != NULL){
            if(l1->val < l2->val){
                p->next = l1;
                p = p->next;
                l1 = l1->next;
            }else{
                p->next = l2;
                p = p->next;
                l2 = l2->next;
            }
        }
        if(l1 != NULL){
            p->next = l1;
        }else if(l2 != NULL){
            p->next = l2;
        }

        return new_head;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(lists.size() == 0){
            return NULL;
        }else if(lists.size() == 1){
            return lists[0];
        }
        auto length = lists.end()-lists.begin();
        auto mid = lists.begin() + (length-1)/2;
        vector<ListNode*>v1(lists.begin(), mid+1);
        vector<ListNode*>v2(mid+1, lists.end());
        //归并排序的思想
        ListNode *l1 = mergeKLists(v1);
        ListNode *l2 = mergeKLists(v2);

        return mergeTwoLists(l1, l2);
    }
};