Question

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

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

/**
 * 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;
  }
}

# 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