23.merge-k-sorted-lists

Statement

简体中文English

Metadata

Link: 合并K个升序链表
Difficulty: Hard
Tag: 链表 分治 堆（优先队列） 归并排序

给你一个链表数组，每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中，返回合并后的链表。

示例 1：

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

示例 2：

输入：lists = []
输出：[]

示例 3：

输入：lists = [[]]
输出：[]

提示：

k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按升序排列
lists[i].length 的总和不超过 10^4

Metadata

Link: Merge k Sorted Lists
Difficulty: Hard
Tag: Linked List Divide and Conquer Heap (Priority Queue) Merge Sort

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

Constraints:

k == lists.length
0 <= k <= 10⁴
0 <= lists[i].length <= 500
-10⁴ <= lists[i][j] <= 10⁴
lists[i] is sorted in ascending order.
The sum of lists[i].length will not exceed 10⁴.

Solution

C++

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

#ifdef LOCAL

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) {}
};

#endif
/**
 * 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) {}
 * };
 */

template <typename T>
struct cmp : public binary_function<T, T, bool> {
    constexpr bool operator()(const T &l, const T &r) const {
        if (!l) {
            return false;
        }

        if (!r) {
            return true;
        }

        return l->val > r->val;
    }
};

class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        priority_queue<ListNode *, vector<ListNode *>, cmp<ListNode *>> pq;

        auto rt = new ListNode();
        auto res = rt;

        for (const auto p : lists) {
            pq.push(p);
        }

        while (!pq.empty()) {
            auto p = pq.top();
            pq.pop();

            if (!p) {
                continue;
            }

            rt->next = new ListNode();
            rt = rt->next;
            rt->val = p->val;

            pq.push(p->next);
        }

        return res->next;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: October 11, 2023