Merge k Sorted Lists

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

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

Solution: 這題可以參考歸併排序演算法的合併過程。採用的是分治的思想。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
 

struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {
    if (listsSize == 0) return NULL;
    if (listsSize == 1) return *lists;

    struct ListNode *pNode1=mergeKLists(lists, listsSize/2);
    struct ListNode *pNode2=mergeKLists(lists+listsSize/2, listsSize-listsSize/2);
    struct ListNode *
 
pNewListRoot,*pNewList;
    if (pNode1 && pNode2 && pNode1->val > pNode2->val) {
        pNewList = pNode2;
        pNode2 = pNode2->next;
    } else if (pNode1) {
        pNewList = pNode1;
        pNode1 = pNode1->next;
    } else if (pNode2) {
        pNewList = pNode2;
 

        pNode2 = pNode2->next;
    } else {
        return NULL;
    }
    pNewListRoot = pNewList;
    while (pNode1 || pNode2) {
        if (pNode1 == NULL) {
            pNewList->next = pNode2;
            pNode2 = pNode2->next;
            pNewList = pNewList->next;
            continue;
        }
        if (pNode2 == NULL) {
            pNewList->next = pNode1;
            pNode1 = pNode1->next;
            pNewList = pNewList->next;
            continue;
        }
        if (pNode1->val > pNode2->val) {
            pNewList->next = pNode2;
            pNode2 = pNode2->next;
        } else {
            pNewList->next = pNode1;
            pNode1 = pNode1->next;
        }
        pNewList = pNewList->next;
    }
    return pNewListRoot;
}