LeetCode Merge K Sorted Lists

class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {   
        ListNode* merged = NULL;
        for (int i=0; i<lists.size(); i++) {
            merged = merge_list(merged, lists[i]);
        }
        return merged;
    }

    // helper function to merge two sorted lists
    ListNode* merge_list(ListNode* l1, ListNode* l2) {
        ListNode* head = NULL;
        ListNode* tail = NULL;
        while (l1 != NULL && l2 != NULL) {
            ListNode* node = NULL;
            if (l1->val <= l2->val) {
                node = l1;
                l1 = l1->next;
            } else {
                node = l2;
                l2 = l2->next;
            }
            if (head == NULL) {
                head = node;
                tail = node;
            } else {
                tail->next = node;
                tail = node;
            }
        }
        if (l1 != NULL) {
            if (head == NULL) {
                head = l1;
            } else {
                tail->next = l1;
            }
        }
        if (l2 != NULL) {
            if (head == NULL) {
                head = l2;
            } else {
                tail->next = l2;
            }
        }
        return head;
    }
};

 第二轮：

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

class MyCmp {
public:
    bool operator() (const ListNode* a, const ListNode* b) {
        return a->val > b->val;
    }
};
 
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {

        priority_queue<ListNode*, vector<ListNode*>, MyCmp> heads;
        
        for (int i=0; i<lists.size(); i++) {
            if (lists[i] == NULL) {
                continue;
            }
            heads.push(lists[i]);
        }
        ListNode* head = NULL;
        ListNode* pre = NULL;
        ListNode* cur = NULL;
        
        while (!heads.empty()) {
            cur = heads.top();
            heads.pop();
            if (head == NULL) {
                head = cur;
            } else {
                pre->next = cur;
            }
            if (cur->next != NULL) {
                heads.push(cur->next);
            }
            pre = cur;
        }
        if (pre != NULL) {
            pre->next = NULL;
        }
        return head;
    }
};

 复杂度mnlog(n), m为链表平均长度，n为链表条数。

看了题解也可以使用分治思想分别合并各条链表。时间复杂度和用堆排序的一致，但是空间可以减少到O(1)

/**
 * 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) {
        int len = lists.size();
        if (len < 1) {
            return NULL;
        }
        int end = len;
        int wpos= 0;
        while (end > 1) {
            for (int i=0; i<end; i+=2) {
                if (i+1 >= end) {
                    lists[wpos++] = lists[i];
                    break;
                }
                lists[wpos++] = mergeLists(lists[i], lists[i+1]);
            }
            end = wpos;
            wpos= 0;
        }
        return lists[0];
    }
    
    ListNode* mergeLists(ListNode* ha, ListNode* hb) {
        ListNode holder(0);
        ListNode* prev = &holder;
        while (ha != NULL && hb != NULL) {
            if (ha->val <= hb->val) {
                prev->next = ha;
                ha = ha->next;
            } else {
                prev->next = hb;
                hb = hb->next;
            }
            prev = prev->next;
        }
        if (ha != NULL) {
            prev->next = ha;
        } 
        if (hb != NULL) {
            prev->next = hb;
        }
        return holder.next;
    }
};