[LeetCode] 23. Merge k Sorted Lists 合并k个有序链表

Posted 2020-10-26 轻风舞动

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

与21. Merge Two Sorted Lists的拓展，这道题要合并k个有序链表。

解法1: 两两合并。1和2合并，结果在和3合并，以此类推，直到结束。时间复杂度为：2n + 3n + ... + kn = [(k+1)*k/2-1]*n = O(nk^2)，空间复杂度为O(1)

解法2: 分治法Divide and Conquer，也就是二分法。每次将所有的list两两之间合并，直到所有list合并成一个。时间复杂度：2n * k/2 + 4n * k/4 + ... + (2^x)n * k/(2^x) = O(nklogk)，如果用迭代空间复杂度为O(1)，用递归则空间复杂度为O(logk)。

解法3: 最小堆MinHeap/priority queue，取每个Linked List的最小节点放入一个heap中，排序成最小堆，取出堆顶最小的元素，放入合并的list中，然后将该节点在其list中的下一个节点插入heap，循环上面步骤，以此类推直到全部节点都经过heap。由于heap的大小为始终为k，而每次插入的复杂度是logk，一共插入了nk个节点。时间复杂度为O(nklogk)，空间复杂度为O(k)。

Python: Merge two by two, Time: O(nk^2), Space: O(1)

class Solution(object):
    def mergeKLists(self, lists):
        def mergeTwoLists(l1, l2):
            curr = dummy = ListNode(0)
            while l1 and l2:
                if l1.val < l2.val:
                    curr.next = l1
                    l1 = l1.next
                else:
                    curr.next = l2
                    l2 = l2.next
                curr = curr.next
            curr.next = l1 or l2
            return dummy.next

        if not lists:
            return None
        left, right = 0, len(lists) - 1;
        while right > 0:
            if left >= right:
                left = 0
            else:
                lists[left] = mergeTwoLists(lists[left], lists[right])
                left += 1
                right -= 1
        return lists[0]

Python: Divide and conquer, Time: O(nklogk), Space: O(logk)

class Solution:
    def mergeKLists(self, lists):
        def mergeTwoLists(l1, l2):
            curr = dummy = ListNode(0)
            while l1 and l2:
                if l1.val < l2.val:
                    curr.next = l1
                    l1 = l1.next
                else:
                    curr.next = l2
                    l2 = l2.next
                curr = curr.next
            curr.next = l1 or l2
            return dummy.next
    
        def mergeKListsHelper(lists, begin, end):
            if begin > end:
                return None
            if begin == end:
                return lists[begin]
            return mergeTwoLists(mergeKListsHelper(lists, begin, (begin + end) / 2), \\
                                 mergeKListsHelper(lists, (begin + end) / 2 + 1, end))
   
        return mergeKListsHelper(lists, 0, len(lists) - 1)

Python: Heap, Time: O(nklogk), Space: O(k)

import heapq
class Solution:
    def mergeKLists(self, lists):
        dummy = ListNode(0)
        current = dummy
        
        heap = []
        for sorted_list in lists:
            if sorted_list:
                heapq.heappush(heap, (sorted_list.val, sorted_list))
                
        while heap:
            smallest = heapq.heappop(heap)[1]
            current.next = smallest
            current = current.next
            if smallest.next:
                heapq.heappush(heap, (smallest.next.val, smallest.next))
                
        return dummy.next　　

C++: Merge two by two

class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        ListNode *ret = NULL;
        for(int i=0; i<lists.size(); i++) 
            ret = merge2Lists(ret, lists[i]);
        return ret;
    }
    
    ListNode* merge2Lists(ListNode *h1, ListNode *h2) {
        ListNode *dummy = new ListNode(0), *tail = dummy;
        while(h1 && h2) {
            if(h1->val<=h2->val) {
                tail->next = h1;
                h1 = h1->next;
            }
            else {
                tail->next = h2;
                h2 = h2->next;
            }
            tail = tail->next;
        }
        tail->next = h1 ? h1 : h2;
        return dummy->next;
    }
};

C++: Divide and Conque

class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        if(lists.empty()) return NULL;
        int end = lists.size()-1;
        while(end>0) {
            int begin = 0;
            while(begin<end) {
                lists[begin] = merge2Lists(lists[begin], lists[end]);
                begin++;
                end--;
            }
        }
        return lists[0];
    }
    
    ListNode* merge2Lists(ListNode *h1, ListNode *h2) {
        ListNode *dummy = new ListNode(0), *tail = dummy;
        while(h1 && h2) {
            if(h1->val<=h2->val) {
                tail->next = h1;
                h1 = h1->next;
            }
            else {
                tail->next = h2;
                h2 = h2->next;
            }
            tail = tail->next;
        }
        tail->next = h1 ? h1 : h2;
        return dummy->next;
    }
};

C++: priority queque

class Solution {
public:
    struct compNode {
        bool operator()(ListNode *p, ListNode *q) const {
            return p->val>q->val;
        }  
    };

    ListNode *mergeKLists(vector<ListNode *> &lists) {
        priority_queue<ListNode*, vector<ListNode*>, compNode> pq;
        ListNode *dummy = new ListNode(0), *tail = dummy;
        
        for(int i=0; i<lists.size(); i++) 
            if(lists[i]) pq.push(lists[i]);
            
        while(!pq.empty()) {
            tail->next = pq.top();
            tail = tail->next;
            pq.pop();
            if(tail->next) pq.push(tail->next);
        }
        
        return dummy->next;
    }
};　　

　　　

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[LeetCode] 21. Merge Two Sorted Lists 合并有序链表

[LeetCode] 88. Merge Sorted Array 合并有序数组

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