2023年8月12日每日一题--23. 合并 K 个升序链表

leetcode链接：23. 合并 K 个升序链表

题目分析

方案一

暴力解决

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode head = new ListNode();
        ListNode node = head;

        int index = -1;
        int tempNum = Integer.MAX_VALUE;

        for (;;){
            index = -1;
            tempNum = Integer.MAX_VALUE;
            for (int i = 0; i <lists.length; i++) {
                if (lists[i] != null && lists[i].val < tempNum) {
                    tempNum = lists[i].val;
                    index = i;
                }
            }
            if (index == -1) break;
            node.next = lists[index];
            node = node.next;
            lists[index] = lists[index].next;
        }
        return head.next;
    }
}

结果

解答成功:
执行耗时:215 ms,击败了5.06% 的Java用户
内存消耗:42.1 MB,击败了81.03% 的Java用户

分析

时间复杂度：
O( n * 数组中的节点数 )

空间复杂度：
O( 1 )

方案二

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {

        PriorityQueue<Integer[]> queue = new PriorityQueue<>((o1, o2) -> o1[0].compareTo(o2[0]));
        ListNode head = new ListNode();
        ListNode node = head;


        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null)
                queue.offer(new Integer[]{lists[i].val, i});
        }

        while (!queue.isEmpty()) {
            Integer[] poll = queue.poll();

            node.next = lists[poll[1]];
            node = node.next;
            lists[poll[1]] = lists[poll[1]].next;

            if (lists[poll[1]] != null) {
                queue.offer(new Integer[]{lists[poll[1]].val, poll[1]});
            }
        }

        return head.next;
    }
}

结果

解答成功:
执行耗时:5 ms,击败了39.24% 的Java用户
内存消耗:43.3 MB,击败了5.00% 的Java用户

分析

时间复杂度：
O( 数组中的节点数 )

空间复杂度：
O( n )

官方题解

https://leetcode.cn/problems/merge-k-sorted-lists/solutions/219756/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/

前置知识：合并两个有序链表

public ListNode mergeTwoLists(ListNode a, ListNode b) {
    if (a == null || b == null) {
        return a != null ? a : b;
    }
    ListNode head = new ListNode(0);
    ListNode tail = head, aPtr = a, bPtr = b;
    while (aPtr != null && bPtr != null) {
        if (aPtr.val < bPtr.val) {
            tail.next = aPtr;
            aPtr = aPtr.next;
        } else {
            tail.next = bPtr;
            bPtr = bPtr.next;
        }
        tail = tail.next;
    }
    tail.next = (aPtr != null ? aPtr : bPtr);
    return head.next;
}

作者：力扣官方题解
链接：https://leetcode.cn/problems/merge-k-sorted-lists/solutions/219756/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

方法一：顺序合并

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode ans = null;
        for (int i = 0; i < lists.length; ++i) {
            ans = mergeTwoLists(ans, lists[i]);
        }
        return ans;
    }

    public ListNode mergeTwoLists(ListNode a, ListNode b) {
        if (a == null || b == null) {
            return a != null ? a : b;
        }
        ListNode head = new ListNode(0);
        ListNode tail = head, aPtr = a, bPtr = b;
        while (aPtr != null && bPtr != null) {
            if (aPtr.val < bPtr.val) {
                tail.next = aPtr;
                aPtr = aPtr.next;
            } else {
                tail.next = bPtr;
                bPtr = bPtr.next;
            }
            tail = tail.next;
        }
        tail.next = (aPtr != null ? aPtr : bPtr);
        return head.next;
    }
}

作者：力扣官方题解
链接：https://leetcode.cn/problems/merge-k-sorted-lists/solutions/219756/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

方法二：分治合并

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return merge(lists, 0, lists.length - 1);
    }

    public ListNode merge(ListNode[] lists, int l, int r) {
        if (l == r) {
            return lists[l];
        }
        if (l > r) {
            return null;
        }
        int mid = (l + r) >> 1;
        return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));
    }

    public ListNode mergeTwoLists(ListNode a, ListNode b) {
        if (a == null || b == null) {
            return a != null ? a : b;
        }
        ListNode head = new ListNode(0);
        ListNode tail = head, aPtr = a, bPtr = b;
        while (aPtr != null && bPtr != null) {
            if (aPtr.val < bPtr.val) {
                tail.next = aPtr;
                aPtr = aPtr.next;
            } else {
                tail.next = bPtr;
                bPtr = bPtr.next;
            }
            tail = tail.next;
        }
        tail.next = (aPtr != null ? aPtr : bPtr);
        return head.next;
    }
}

作者：力扣官方题解
链接：https://leetcode.cn/problems/merge-k-sorted-lists/solutions/219756/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

方法三：使用优先队列合并

class Solution {
    class Status implements Comparable<Status> {
        int val;
        ListNode ptr;

        Status(int val, ListNode ptr) {
            this.val = val;
            this.ptr = ptr;
        }

        public int compareTo(Status status2) {
            return this.val - status2.val;
        }
    }

    PriorityQueue<Status> queue = new PriorityQueue<Status>();

    public ListNode mergeKLists(ListNode[] lists) {
        for (ListNode node: lists) {
            if (node != null) {
                queue.offer(new Status(node.val, node));
            }
        }
        ListNode head = new ListNode(0);
        ListNode tail = head;
        while (!queue.isEmpty()) {
            Status f = queue.poll();
            tail.next = f.ptr;
            tail = tail.next;
            if (f.ptr.next != null) {
                queue.offer(new Status(f.ptr.next.val, f.ptr.next));
            }
        }
        return head.next;
    }
}

作者：力扣官方题解
链接：https://leetcode.cn/problems/merge-k-sorted-lists/solutions/219756/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。