算法leetcode｜23. 合并K个升序链表（rust重拳出击）

Posted 2023-01-17 二当家的白帽子

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文章目录

• 23. 合并K个升序链表：
• • 样例 1：
  • 样例 2：
  • 样例 3：
  • 提示：
• 分析：
• 题解：
• • rust
  • go
  • c++
  • c
  • python
  • java

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23. 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。

样例 1：

输入：
	lists = [[1,4,5],[1,3,4],[2,6]]
	
输出：
	[1,1,2,3,4,4,5,6]
	
解释：
	链表数组如下：
	[
	  1->4->5,
	  1->3->4,
	  2->6
	]
	将它们合并到一个有序链表中得到。
	1->1->2->3->4->4->5->6

样例 2：

输入：
	lists = []
	
输出：
	[]

样例 3：

输入：
	lists = [[]]
	
输出：
	[]

提示：

• k == lists.length
• 0 <= k <= 10⁴
• 0 <= lists[i].length <= 500
• -10⁴ <= lists[i][j] <= 10⁴
• lists[i] 按 升序 排列
• lists[i].length 的总和不超过 10⁴

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分析：

• 面对这道算法题目，二当家的陷入了沉思。
• 合并有序链表或者合并有序数组非常像归并排序的合并部分。
• 递归和迭代都可以，通常递归更加简单直观，迭代更加高效。
• 可以按顺序合并，但是这样每次只能合并掉一个链表，而用类似归并排序的分治方式则可以每次减少一半数量的链表。

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题解：

rust

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode 
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// 
//
// impl ListNode 
//   #[inline]
//   fn new(val: i32) -> Self 
//     ListNode 
//       next: None,
//       val
//     
//   
// 
impl Solution 
    pub fn merge_k_lists(mut lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> 
        fn merge(lists: &mut Vec<Option<Box<ListNode>>>, l: usize, r: usize) -> Option<Box<ListNode>> 
            fn merge_two_lists(list1: Option<Box<ListNode>>, list2: Option<Box<ListNode>>) -> Option<Box<ListNode>> 
                match (list1, list2) 
                    (None, None) => None,
                    (None, r) => r,
                    (l, None) => l,
                    (Some(mut l), Some(mut r)) => 
                        if l.val <= r.val 
                            l.next = merge_two_lists(l.next, Some(r));
                            Some(l)
                         else 
                            r.next = merge_two_lists(Some(l), r.next);
                            Some(r)
                        
                    
                
            

            if l == r 
                return lists[l].take();
            
            if l > r 
                return None;
            
            let mid = (l + r) >> 1;
            return merge_two_lists(merge(lists, l, mid), merge(lists, mid + 1, r));
        

        let l = lists.len();
        if l == 0 
            return None;
        
        merge(&mut lists, 0, l - 1)
    

---

go

/**
 * Definition for singly-linked list.
 * type ListNode struct 
 *     Val int
 *     Next *ListNode
 * 
 */
func mergeKLists(lists []*ListNode) *ListNode 
    var merge func(int, int) *ListNode
	merge = func(l int, r int) *ListNode 
		var mergeTwoLists func(*ListNode, *ListNode) *ListNode
		mergeTwoLists = func(list1 *ListNode, list2 *ListNode) *ListNode 
			if nil == list1 
				return list2
			

			if nil == list2 
				return list1
			

			if list1.Val < list2.Val 
				list1.Next = mergeTwoLists(list1.Next, list2)
				return list1
			 else 
				list2.Next = mergeTwoLists(list1, list2.Next)
				return list2
			
		

		if l == r 
			return lists[l]
		
		if l > r 
			return nil
		
		mid := (l + r) >> 1
		return mergeTwoLists(merge(l, mid), merge(mid+1, r))
	

	return merge(0, len(lists)-1)

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c++

/**
 * Definition for singly-linked list.
 * struct ListNode 
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) 
 *     ListNode(int x) : val(x), next(nullptr) 
 *     ListNode(int x, ListNode *next) : val(x), next(next) 
 * ;
 */
class Solution 
private:
    ListNode *mergeTwoLists(ListNode *list1, ListNode *list2) 
        if (!list1) 
            return list2;
        

        if (!list2) 
            return list1;
        

        if (list1->val < list2->val) 
            list1->next = mergeTwoLists(list1->next, list2);
            return list1;
         else 
            list2->next = mergeTwoLists(list1, list2->next);
            return list2;
        
    

    ListNode *merge(vector<ListNode *> &lists, int l, int r) 
        if (l == r) 
            return lists[l];
        
        if (l > r) 
            return nullptr;
        
        int mid = (l + r) >> 1;
        return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));
    

public:
    ListNode *mergeKLists(vector<ListNode *> &lists) 
        return merge(lists, 0, lists.size() - 1);
    
;

---

c

/**
 * Definition for singly-linked list.
 * struct ListNode 
 *     int val;
 *     struct ListNode *next;
 * ;
 */


struct ListNode *mergeTwoLists(struct ListNode *list1, struct ListNode *list2) 
    if (!list1) 
        return list2;
    

    if (!list2) 
        return list1;
    

    if (list1->val < list2->val) 
        list1->next = mergeTwoLists(list1->next, list2);
        return list1;
     else 
        list2->next = mergeTwoLists(list1, list2->next);
        return list2;
    


struct ListNode *merge(struct 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));


struct ListNode *mergeKLists(struct ListNode **lists, int listsSize) 
    return merge(lists, 0, listsSize - 1);

---

python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        def merge(l: int, r: int) -> Optional[ListNode]:
            def mergeTwoLists(list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
                if list1 is None:
                    return list2
                if list2 is None:
                    return list1
                if list1.val < list2.val:
                    list1.next = mergeTwoLists(list1.next, list2)
                    return list1
                else:
                    list2.next = mergeTwoLists(list1, list2.next)
                    return list2

            if l == r:
                return lists[l]
            if l > r:
                return None
            mid = (l + r) >> 1
            return mergeTwoLists(merge(l, mid), merge(mid + 1, r))

        return merge(0, len(lists) - 1)

---

java

/**
 * 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 list1, ListNode list2) 
        if (list1 == null) 
            return list2;
        

        if (list2 == null) 
            return list1;
        

        if (list1.val < list2.val) 
            list1.next = mergeTwoLists(list1.next, list2);
            return list1;
         else 
            list2.next = mergeTwoLists(list1, list2.next);
            return list2;
        
    

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