LeetCode题解：Merge k Sorted Lists

Merge k Sorted Lists

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

思路：

标准的merge sort使用min(x,y)，这里需要的是min(x1, x2, ... xk)，可以用堆实现。

题解：

/** * 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) {        using node_pair = pair<ListNode*, size_t>;               // Note: By default, C++ will create a max-heap, but we need a min-heap        auto comparison=[](const node_pair& n1, const node_pair& n2) -> bool {            return n1.first->val > n2.first->val;         };               // starting point of the list        ListNode* new_start = new ListNode(0);        ListNode* new_last = new_start;               // Heap, storing the first node of each list        vector<node_pair> v_nodes;        v_nodes.reserve(lists.size());               // initalize the heap        size_t index=0;        for(auto& t : lists)        {            if (t!=nullptr)            {                v_nodes.push_back(node_pair(t, index));                t=t->next;            }            index++;        }        make_heap(begin(v_nodes), end(v_nodes),comparison);               // Heap injector        auto inject_heap=[&v_nodes,&lists,&comparison](const size_t list_id) -> void {            if (lists[list_id]==nullptr)                return;            else            {                auto save = lists[list_id];                lists[list_id]=save->next;                v_nodes.push_back(node_pair(save, list_id));                push_heap(begin(v_nodes), end(v_nodes), comparison);            }        };                // Loop until the heap is empty        while(!v_nodes.empty())        {            pop_heap(begin(v_nodes), end(v_nodes), comparison);            auto t = v_nodes.back();            v_nodes.pop_back();            new_last->next = t.first;            new_last = new_last->next;                       inject_heap(t.second);        }               // drop the head        new_last=new_start;        new_start=new_start->next;        delete new_last;               return new_start;    }};