21. Merge Two Sorted Lists

本文介绍了一种合并两个已排序链表的方法,通过遍历两个链表并比较节点值来保持合并后链表的有序性。提供了Python实现示例。

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题目大意:
有两个已经排好序的链表,现在要合并这两个链表,同时保证大小顺序不变,返回合并后的链表,如:1->3->5和1->4合并后为:1->1->3->4->5

思路:
令i,j分别为两个链表的起始节点,选择值较小的那个节点,加入到返回链表中,此处i指向的节点的值为1,j指向节点的值为1,两个值相等,(相等时默认加入i),此处把 i 指向的节点加入到返回链表中,随后 i 后移一位。
之后,i指向的节点的值为3,j指向的节点的值为1,1 < 3,所以把j指向的节点加入到返回链表中。
之后同上。。
最后,当j == null时,把i及i之后的节点链表直接加入到返回链表中即可。
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Python代码:

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        i = l1
        j = l2
        r = ListNode(-1)
        h = r
        while i != None and j != None:
            if i.val <= j.val:
                r.next = i
                i = i.next
            else:
                r.next = j
                j = j.next
            r = r.next

        while i != None:
            r.next = i
            i = i.next
            r = r.next
        while j != None:
            r.next = j
            j = j.next
            r = r.next

        return h.next
To merge k sorted linked lists, one approach is to repeatedly merge two of the linked lists until all k lists have been merged into one. We can use a priority queue to keep track of the minimum element across all k linked lists at any given time. Here's the code to implement this idea: ``` struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; // Custom comparator for the priority queue struct CompareNode { bool operator()(const ListNode* node1, const ListNode* node2) const { return node1->val > node2->val; } }; ListNode* mergeKLists(vector<ListNode*>& lists) { priority_queue<ListNode*, vector<ListNode*>, CompareNode> pq; for (ListNode* list : lists) { if (list) { pq.push(list); } } ListNode* dummy = new ListNode(-1); ListNode* curr = dummy; while (!pq.empty()) { ListNode* node = pq.top(); pq.pop(); curr->next = node; curr = curr->next; if (node->next) { pq.push(node->next); } } return dummy->next; } ``` We start by initializing a priority queue with all the head nodes of the k linked lists. We use a custom comparator that compares the values of two nodes and returns true if the first node's value is less than the second node's value. We then create a dummy node to serve as the head of the merged linked list, and a current node to keep track of the last node in the merged linked list. We repeatedly pop the minimum node from the priority queue and append it to the merged linked list. If the popped node has a next node, we push it onto the priority queue. Once the priority queue is empty, we return the head of the merged linked list. Note that this implementation has a time complexity of O(n log k), where n is the total number of nodes across all k linked lists, and a space complexity of O(k).
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