LeetCode 23. 合并 K 个升序链表

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题目描述
思路分析
参考代码

题目描述

✅ 23. 合并 K 个升序链表

思路分析

✅ 解法一：最小堆（推荐）

✅ 解法二：分治合并

✅ 解法三：顺序合并

参考代码

type ListHeap []*ListNode

func (h ListHeap) Len() int            { return len(h) }
func (h ListHeap) Less(i, j int) bool  { return h[i].Val < h[j].Val }
func (h ListHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
func (h *ListHeap) Push(x interface{}) { *h = append(*h, x.(*ListNode)) }
func (h *ListHeap) Pop() interface{} {
	old := *h
	n := len(old)
	node := old[n-1]
	*h = old[:n-1]
	return node
}

func mergeKLists(lists []*ListNode) *ListNode {
	h := &ListHeap{}
	heap.Init(h)

	for _, node := range lists {
		if node != nil {
			heap.Push(h, node)
		}
	}

	dummy := &ListNode{}
	cur := dummy

	for h.Len() > 0 {
		node := heap.Pop(h).(*ListNode)
		cur.Next = node
		cur = cur.Next
		if node.Next != nil {
			heap.Push(h, node.Next)
		}
	}

	return dummy.Next
}

func mergeKLists(lists []*ListNode) *ListNode {
	if len(lists) == 0 {
		return nil
	} else if len(lists) == 1 {
		return lists[0]
	}
	mid := len(lists) / 2
	l1 := mergeKLists(lists[:mid])
	l2 := mergeKLists(lists[mid:])
	return mergeTwoLists(l1, l2)
}

func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	} else if l2 == nil {
		return l1
	}
	dummy := &ListNode{}
	cur := dummy
	for l1 != nil && l2 != nil {
		if l1.Val < l2.Val {
			cur.Next = l1
			l1 = l1.Next
		} else {
			cur.Next = l2
			l2 = l2.Next
		}
		cur = cur.Next
	}
	if l1 != nil {
		cur.Next = l1
	} else {
		cur.Next = l2
	}
	return dummy.Next
}

func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	} else if l2 == nil {
		return l1
	}
	if l1.Val < l2.Val {
		l1.Next = mergeTwoLists(l1.Next, l2)
		return l1
	} else {
		l2.Next = mergeTwoLists(l1, l2.Next)
		return l2
	}
}

// 合并 K 个升序链表
func mergeKLists(lists []*ListNode) *ListNode {
	if len(lists) == 0 {
		return nil
	}

	// 从第一个链表开始
	mergedList := lists[0]

	// 依次合并每个链表
	for i := 1; i < len(lists); i++ {
		mergedList = mergeTwoLists(mergedList, lists[i])
	}

	return mergedList
}

// 合并两个有序链表
func mergeTwoLists(l1, l2 *ListNode) *ListNode {
	dummy := &ListNode{}
	cur := dummy

	for l1 != nil && l2 != nil {
		if l1.Val < l2.Val {
			cur.Next = l1
			l1 = l1.Next
		} else {
			cur.Next = l2
			l2 = l2.Next
		}
		cur = cur.Next
	}

	if l1 != nil {
		cur.Next = l1
	} else {
		cur.Next = l2
	}

	return dummy.Next
}

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本文作者: HGNULB

本文链接: https://hgnulb.github.io/blog/2025/59897639