23. 合并K个升序链表

顺序合并

/**
 * 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) {
        ListNode res = null;
        for (int i = 0; i < lists.length; i++) {
            res = mergeTwoLists(res, lists[i]);
        }
        return res;
    }

    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        p.next = l1 != null ? l1 : l2;
        return dummy.next;
    }
}

• 时间复杂度 O(k²n)
• 空间复杂度 O(1)

分治合并

/**
 * 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) {
        return merge(lists, 0, lists.length - 1);
    }

    private 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));
    }

    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        p.next = l1 != null ? l1 : l2;
        return dummy.next;
    }
}

• 时间复杂度 O(kn * logk)
• 空间复杂度 O(logk)

K个指针比较，选取最小值

/**
 * 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) {
        int k = lists.length;
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (true) {
            // minNode记录k个节点中值最小的节点
            ListNode minNode = null;
            // minIndex记录最小节点在数组中的索引
            int minIndex = -1;
            // 比较K个节点
            for (int i = 0; i < k; i++) {
                if (lists[i] == null) {
                    continue;
                }
                if (minNode == null || lists[i].val < minNode.val) {
                    minNode = lists[i];
                    minIndex = i;
                }
            }
            // 如果最小节点为空，则说明已经遍历完所有链表
            if (minIndex == -1) {
                break;
            }
            // 移动结果链表
            p.next = minNode;
            p = p.next;
            lists[minIndex] = lists[minIndex].next;
        }
        return dummy.next; 
    }   
}

• 时间复杂度 O(k²n)
• 空间复杂度 O(1)

优先级队列构建最小堆

/**
 * 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) {
        int k = lists.length;
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        Queue<ListNode> heap = new PriorityQueue<>((v1, v2) -> v1.val - v2.val);
        for (ListNode node: lists) {
            if (node != null) {
                heap.offer(node);
            }
        }
        while (!heap.isEmpty()) {
            ListNode minNode = heap.poll();
            p.next = minNode;
            p = p.next;
            if (minNode.next != null) {
                heap.offer(minNode.next);
            }
        }
        return dummy.next; 
    }  
}

• 时间复杂度 O(kn * logk)
• 空间复杂度 O(k)