c++学习笔记：有序链表的合并

/*
LeetCode_0023
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
*/
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

1. 以递归的方式

ListNode* mergeTwo(ListNode* l1, ListNode* l2) {
        if (l1 == NULL)
            return l2;
        if (l2 == NULL)
            return l1;
        if (l1->val <= l2->val) {
            l1->next = mergeTwo(l1->next, l2);
            return l1;
        }
        if (l1->val > l2->val) {
            l2->next = mergeTwo(l1, l2->next);
            return l2;
        }
    }

2. 直接合并

    ListNode* mergeTwo(ListNode* l1, ListNode* l2) {
        ListNode dummy(-1);
        ListNode* result = &dummy;
        while (l1&&l2) {
            if (l1->val <= l2->val) {
                result->next = l1;
                result = result->next;
                l1 = l1->next;
            }
            else if (l1->val > l2->val) {
                result->next = l2;
                result = result->next;
                l2 = l2->next;
            }
        }
        while (l1)
            result->next = l1;
        while (l2)
            result->next = l2;
        return dummy.next;
    }

3. 合并k个有序链表

ListNode *mergeKLists(vector<ListNode *> &lists) {
    if (lists.empty()) {
        return nullptr;
    }
    while (lists.size() > 1) {
        lists.push_back(mergeTwoLists(lists[0], lists[1]));
        lists.erase(lists.begin());
        lists.erase(lists.begin());
    }
    return lists.front();
}
ListNode *mergeKLists(vector<ListNode *> &lists) {
    if (lists.empty()) return NULL;
        int len = lists.size();
        while (len > 1) 
        {
            for (int i = 0; i < len / 2; ++i) 
            {
                lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);
            }
            len = (len + 1) / 2;
        }
        return lists.front();
}