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剑指 Offer II 078. 合并排序链表

题目描述

给定一个链表数组，每个链表都已经按升序排列。

请将所有链表合并到一个升序链表中，返回合并后的链表。

 

示例 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

示例 2：

输入：lists = []
输出：[]

示例 3：

输入：lists = [[]]
输出：[]

 

提示：

  • k == lists.length
  • 0 <= k <= 10^4
  • 0 <= lists[i].length <= 500
  • -10^4 <= lists[i][j] <= 10^4
  • lists[i] 按 升序 排列
  • lists[i].length 的总和不超过 10^4

 

注意：本题与主站 23 题相同： https://leetcode.cn/problems/merge-k-sorted-lists/

解法

方法一

Python3

# 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[ListNode]) -> ListNode:
        n = len(lists)
        if n == 0:
            return None
        for i in range(n - 1):
            lists[i + 1] = self.mergeTwoLists(lists[i], lists[i + 1])
        return lists[-1]

    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode()
        cur = dummy
        while l1 and l2:
            if l1.val <= l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        cur.next = l1 or l2
        return dummy.next

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) {
        int n = lists.length;
        if (n == 0) {
            return null;
        }
        for (int i = 0; i < n - 1; ++i) {
            lists[i + 1] = mergeLists(lists[i], lists[i + 1]);
        }
        return lists[n - 1];
    }

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

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 {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int n = lists.size();
        if (n == 0) return nullptr;
        for (int i = 1; i < n; ++i) lists[i] = mergeTwoLists(lists[i - 1], lists[i]);
        return lists[n - 1];
    }

private:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode();
        ListNode* cur = dummy;
        while (l1 && l2) {
            if (l1->val <= l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        cur->next = l1 ? l1 : l2;
        return dummy->next;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
	n := len(lists)
	if n == 0 {
		return nil
	}
	for i := 1; i < n; i++ {
		lists[i] = mergeTwoLists(lists[i-1], lists[i])
	}
	return lists[n-1]
}

func mergeTwoLists(l1 *ListNode, 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 if l2 != nil {
		cur.Next = l2
	}
	return dummy.Next
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function (lists) {
    const n = lists.length;
    if (n == 0) {
        return null;
    }
    for (let i = 1; i < n; ++i) {
        lists[i] = mergeTwoLists(lists[i - 1], lists[i]);
    }
    return lists[n - 1];
};

function mergeTwoLists(l1, l2) {
    const dummy = new ListNode();
    let cur = dummy;
    while (l1 && l2) {
        if (l1.val <= l2.val) {
            cur.next = l1;
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next;
        }
        cur = cur.next;
    }
    cur.next = l1 || l2;
    return dummy.next;
}

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode MergeKLists(ListNode[] lists) {
        int n = lists.Length;
        if (n == 0) {
            return null;
        }
        for (int i = 1; i < n; ++i) {
            lists[i] = MergeTwoLists(lists[i - 1], lists[i]);
        }
        return lists[n - 1];
    }

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

Ruby

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode[]} lists
# @return {ListNode}
def merge_k_lists(lists)
    n = lists.length
    i = 1
    while i < n
        lists[i] = merge_two_lists(lists[i - 1], lists[i])
        i += 1
    end
    lists[n - 1]
end

def merge_two_lists(l1, l2)
  dummy = ListNode.new()
  cur = dummy
  while l1 && l2
      if l1.val <= l2.val
          cur.next = l1
          l1 = l1.next
      else
          cur.next = l2
          l2 = l2.next
      end
      cur = cur.next
  end
  cur.next = l1 || l2
  dummy.next
end

Swift

/** class ListNode {
*    var val: Int
*    var next: ListNode?
*    init() { self.val = 0; self.next = nil }
*    init(_ val: Int) { self.val = val; self.next = nil }
*    init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next }
* }
*/

class Solution {
    func mergeKLists(_ lists: [ListNode?]) -> ListNode? {
        let n = lists.count
        if n == 0 {
            return nil
        }

        var mergedList: ListNode? = lists[0]
        for i in 1..<n {
            mergedList = mergeLists(mergedList, lists[i])
        }
        return mergedList
    }

    private func mergeLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        let dummy = ListNode()
        var cur = dummy
        var l1 = l1
        var l2 = l2

        while let node1 = l1, let node2 = l2 {
            if node1.val <= node2.val {
                cur.next = node1
                l1 = node1.next
            } else {
                cur.next = node2
                l2 = node2.next
            }
            cur = cur.next!
        }
        cur.next = l1 ?? l2
        return dummy.next
    }
}