[LeetCode] 23. Merge k Sorted Lists

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.

合并 K 个升序链表。

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。

这道题有两种做法，时间空间复杂度一样，一个是用priority queue 解决，一个是用 merge sort 的思路解决。

首先是用 priority queue 的思路。因为JS实现 PQ 太过麻烦，所以 JS 实现我给出次优解，思路是 merge sort。

时间O(nlogk) - k 是链表的数量，最多有 k 个 node 在 pq 中，每次放一个新的 node 进去就要排序一次；所有的 node 都会被 pq 筛一遍，所以再乘以 n

空间O(n)

JavaScript实现

/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function (lists) {
    return divide(lists, 0, lists.length - 1);
};

var divide = function (lists, start, end) {
    if (start === end) {
        return lists[start];
    } else if (start < end) {
        const mid = parseInt(start + (end - start) / 2);
        const left = divide(lists, start, mid);
        const right = divide(lists, mid + 1, end);
        return merge(left, right);
    } else {
        return null;
    }
}

var merge = function (left, right) {
    if (!left) {
        return right;
    } else if (!right) {
        return left;
    } else if (left.val < right.val) {
        left.next = merge(left.next, right);
        return left;
    } else {
        right.next = merge(left, right.next);
        return right;
    }
}

 

Java实现 - priority queue

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        // corner case
        if (lists == null || lists.length == 0) {
            return null;
        }

        // normal case
        PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, (a, b) -> a.val - b.val);
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        for (ListNode list : lists) {
            // 注意sublist有可能是空的
            if (list != null) {
                queue.add(list);
            }
        }

        while (!queue.isEmpty()) {
            cur.next = queue.poll();
            cur = cur.next;
            if (cur.next != null) {
                queue.add(cur.next);
            }
        }
        return dummy.next;
    }
}

 

如下是Java实现的 merge sort 的思路。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        // corner case
        if (lists == null || lists.length == 0) {
            return null;
        }
        return sort(lists, 0, lists.length - 1);
    }
    
    private ListNode sort(ListNode[] lists, int lo, int hi) {
        if (lo >= hi) {
            return lists[lo];
        }
        int mid = lo + (hi - lo) / 2;
        ListNode l1 = sort(lists, lo, mid);
        ListNode l2 = sort(lists, mid + 1, hi);
        return merge(l1, l2);
    }
    
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        if (l1 == null) {
            cur.next = l2;
        }
        if (l2 == null) {
            cur.next = l1;
        }
        return dummy.next;
    }
}

 

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