leetcode 23. Merge k Sorted Lists

来源:互联网 发布:淘宝网耐克男特价 编辑:程序博客网 时间:2024/06/02 03:47

相关问题

Discription

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

思路

最小堆
该题为二路归并的变种,多路归并。使用最小堆结构在log(k)时间复杂度内找到当前最小元素。

时间复杂度:nlog(k)
空间复杂度:O(n)

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */bool cmp(const ListNode *a, const ListNode *b){    return a->val > b->val;}class Solution {public:    ListNode* mergeKLists(vector<ListNode*>& lists) {        if (lists.empty())            return NULL;        vector<ListNode*> heap;        ListNode *root, *pNode;        for (int i = 0; i < lists.size(); i++)        {            if (lists[i])                heap.push_back(lists[i]);        }        if (heap.empty())            return NULL;        make_heap(heap.begin(), heap.end(), cmp);        pop_heap(heap.begin(), heap.end(), cmp);        pNode = root = heap.back();        heap.pop_back();        while (!heap.empty())        {            // 加入元素            if (pNode->next)            {                heap.push_back(pNode->next);                push_heap(heap.begin(), heap.end(), cmp);            }            // 取出元素            pop_heap(heap.begin(), heap.end(), cmp);            pNode->next = heap.back();            heap.pop_back();            pNode = pNode->next;        }        return root;    }};