快手一面｜合并 K 个升序链表

输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        auto cmp = [](const ListNode* a, const ListNode* b) {
            return a->val > b->val;
        };
        priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> pq;
        for (auto& head : lists) {
            if (head) {
                pq.push(head);
            }
        }
        ListNode dummy{};
        ListNode* head = &dummy;
        while (!pq.empty()) {
            ListNode* node = pq.top();
            pq.pop();
            head->next = node;
            head = head->next;
            if (node->next)
                pq.push(node->next);
        }
        return dummy.next;
    }
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
        ListNode dummy{};
        ListNode* cur = &dummy;
        while (list1 && list2) {
            if (list1->val < list2->val) {
                cur->next = list1;
                list1 = list1->next;
            } else {
                cur->next = list2;
                list2 = list2->next;
            }
            cur = cur->next;
        }
        cur->next = list1 ? list1 : list2;
        return dummy.next;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists, int l, int r) {
        if (l == r)
            return lists[l];
        if (l > r)
            return nullptr;
        int m = (l + r) >> 1;
        auto left = mergeKLists(lists, l, m);
        auto right = mergeKLists(lists, m + 1, r);
        return mergeTwoLists(left, right);
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        return mergeKLists(lists, 0, lists.size() - 1);
    }
};

class Solution {
public:
    // 向下调整
    void heapify(vector<int>& nums, int i, int n) {
        int largest = i;
        int left = i * 2 + 1;
        int right = i * 2 + 2;
        if (left < n && nums[left] > nums[largest])
            largest = left;
        if (right < n && nums[right] > nums[largest])
            largest = right;
        if (largest != i) {
            swap(nums[largest], nums[i]);
            heapify(nums, largest, n);
        }
    }

    void heapsort(vector<int>& nums) {
        int n = nums.size();
        // i = n / 2 也可, 多判断一次而已
        // 向上初始化构建, 获取数组最大值
        for (int i = n / 2 - 1; i >= 0; i--) {
            heapify(nums, i, n);
        }
        // 最大值不断调整到末尾, 并对新元素 nums[0] 向下进行调整
        for (int i = n - 1; i >= 0; i--) {
            swap(nums[i], nums[0]);
            heapify(nums, 0, i);    // 每次都从 0 开始调整
        }
    }

    vector<int> sortArray(vector<int>& nums) {
        heapsort(nums);
        return nums;
    }
};