Leetcode实战： 23. 合并K个排序链表

题目

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

算法实现：方法一，两两合并

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int length = lists.size();
        if (length == 0) {
            return nullptr;
        }
        else if (length == 1) {
            return lists[0];
        }
        ListNode* p = lists[0];
        for (int i = 1; i < length; i++) {
            p = merge2(p, lists[i]);
        }
        return p;
    }
    ListNode* merge2(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(0);
        ListNode* p = head;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                p->next = l1;
                l1 = l1->next;
            }
            else {
                p->next = l2;
                l2 = l2->next;
            }
            p = p->next;
        }
        while (l1) {
            p->next = l1;
            l1 = l1->next;
            p = p->next;
        }
        while (l2) {
            p->next = l2;
            l2 = l2->next;
            p = p->next;
        }
        return head->next;
    }
};

结果

可以看到，这种方法是比较慢的

改进：如果一个链表为空，直接将另一个链在链表尾部

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int length = lists.size();
        if (length == 0) {
            return nullptr;
        }
        else if (length == 1) {
            return lists[0];
        }
        ListNode* p = lists[0];
        for (int i = 1; i < length; i++) {
            p = merge2(p, lists[i]);
        }
        return p;
    }
    ListNode* merge2(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(0);
        ListNode* p = head;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                p->next = l1;
                l1 = l1->next;
            }
            else {
                p->next = l2;
                l2 = l2->next;
            }
            p = p->next;
        }
        p->next = l1 ? l1 : l2;
        return head->next;
    }
};

可以看到快了很多

方法二：双端队列

class Solution {
public:
	ListNode* mergeKLists(vector<ListNode*>& lists) {
		int length = lists.size();
		if (length == 0) {
			return nullptr;
		}
		else if (length == 1) {
			return lists[0];
		}
		queue<ListNode*> waiting(deque<ListNode*>(lists.begin(), lists.end()));
		while (waiting.size() > 1) {
			ListNode* l1 = waiting.front();
			waiting.pop();
			ListNode* l2 = waiting.front();
			waiting.pop();
			waiting.push(merge2(l1, l2));
		}
		return waiting.front();
	}
	ListNode* merge2(ListNode* l1, ListNode* l2) {
		ListNode* head = new ListNode(0);
		ListNode* p = head;
		while (l1 && l2) {
			if (l1->val < l2->val) {
				p->next = l1;
				l1 = l1->next;
			}
			else {
				p->next = l2;
				l2 = l2->next;
			}
			p = p->next;
		}
		p->next = l1 ? l1 : l2;
		return head->next;
	}
};

结果：