技术求助：如何对单链表按k个节点分组进行排序？

Hey there! Since you already have a working sortedMerge function for full linked list merge sort, we can adapt that logic to solve the grouped sorting problem. Here's a step-by-step breakdown and implementation:

Key Steps to Solve This #

• Split the original list into chunks: Each chunk has up to k nodes. We need to handle the last chunk which might have fewer than k nodes.
• Sort each individual chunk: Use your existing merge sort logic (or even insertion sort, which is efficient for small k) on each chunk.
• Reconnect all sorted chunks: Link the end of one sorted chunk to the start of the next sorted chunk to form the final list.

Implementation Code #

First, let's add helper functions to split the list and handle the grouping, then integrate your existing sortedMerge for sorting chunks.

class LinkedListNode {
    int val;
    LinkedListNode next;
    LinkedListNode(int val) { this.val = val; }
}

// Your existing sortedMerge function (completed)
LinkedListNode sortedMerge(LinkedListNode a, LinkedListNode b) {
    LinkedListNode result = null;
    if (a == null) return b;
    if (b == null) return a;
    
    if (a.val <= b.val) {
        result = a;
        result.next = sortedMerge(a.next, b);
    } else {
        result = b;
        result.next = sortedMerge(a, b.next);
    }
    return result;
}

// Helper to perform merge sort on a single linked list chunk
LinkedListNode mergeSort(LinkedListNode head) {
    if (head == null || head.next == null) return head;
    
    // Split into two halves
    LinkedListNode mid = getMiddle(head);
    LinkedListNode nextOfMid = mid.next;
    mid.next = null;
    
    // Recursively sort both halves
    LinkedListNode left = mergeSort(head);
    LinkedListNode right = mergeSort(nextOfMid);
    
    // Merge the sorted halves
    return sortedMerge(left, right);
}

// Helper to find the middle node of a linked list
LinkedListNode getMiddle(LinkedListNode head) {
    if (head == null) return head;
    LinkedListNode slow = head, fast = head.next;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
    }
    return slow;
}

// Helper to split the list into k-length chunks, sort each, and reconnect
LinkedListNode sortKGroupedLinkedList(LinkedListNode head, int k) {
    if (head == null || k <= 1) return head;
    
    LinkedListNode dummy = new LinkedListNode(0);
    dummy.next = head;
    LinkedListNode prevChunkTail = dummy;
    LinkedListNode current = head;
    
    while (current != null) {
        // Step 1: Find the end of the current k-length chunk
        LinkedListNode chunkHead = current;
        int count = 1;
        while (current.next != null && count < k) {
            current = current.next;
            count++;
        }
        // Save the start of the next chunk
        LinkedListNode nextChunkHead = current.next;
        // Cut off the current chunk from the rest of the list
        current.next = null;
        
        // Step 2: Sort the current chunk
        LinkedListNode sortedChunk = mergeSort(chunkHead);
        
        // Step 3: Link the sorted chunk to the previous part
        prevChunkTail.next = sortedChunk;
        
        // Move prevChunkTail to the end of the sorted chunk
        while (prevChunkTail.next != null) {
            prevChunkTail = prevChunkTail.next;
        }
        
        // Move to the next chunk
        current = nextChunkHead;
    }
    
    return dummy.next;
}

// Test the implementation
public static void main(String[] args) {
    // Build the example list: 4->8->3->1->9->2
    LinkedListNode head = new LinkedListNode(4);
    head.next = new LinkedListNode(8);
    head.next.next = new LinkedListNode(3);
    head.next.next.next = new LinkedListNode(1);
    head.next.next.next.next = new LinkedListNode(9);
    head.next.next.next.next.next = new LinkedListNode(2);
    
    int k = 3;
    LinkedListNode result = sortKGroupedLinkedList(head, k);
    
    // Print the result: 3->4->8->1->2->9
    while (result != null) {
        System.out.print(result.val + "->");
        result = result.next;
    }
    System.out.println("null");
}

Explanation of Key Parts #

1. Chunk Splitting: The loop in sortKGroupedLinkedList iterates through the list, marking the start and end of each k-length chunk. We cut off the chunk from the rest of the list to sort it independently.
2. Chunk Sorting: We reuse your merge sort logic (via mergeSort which calls your sortedMerge) to sort each chunk. If k is small, you could also use insertion sort here for slightly better performance, but merge sort works perfectly too.
3. Reconnecting Chunks: After sorting a chunk, we link it to the end of the previously processed sorted chunks, then move the tail pointer forward to the end of the new sorted chunk.

Edge Cases Handled #

• k = 1: The function returns the original list since no sorting is needed.
• k larger than the list length: The entire list is sorted, just like your full merge sort.
• Empty list: Returns null immediately.

内容的提问来源于stack exchange，提问作者roger_that