LeetCode 23. Merge k Sorted Lists

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

/** * 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) {        heap.resize(lists.size());        int n = 0;        for (int i = 0; i < lists.size(); ++i) {            if (lists[i] != NULL) {                heap[n++] = lists[i];            }        }        if (n == 0) return NULL;        if (n == 1) return heap[0];        BuildMinHeap(heap, n);        ListNode* result = heap[0];        ListNode* cur = result;        while (n > 1) {            if (heap[0]->next != NULL) {                heap[0] = heap[0]->next;            }            else {                heap[0] = heap[--n];            }            MinHeapify(heap, n, 0);            cur->next = heap[0];            cur = cur->next;        }        return result;    }private:    void MinHeapify(vector<ListNode*>& heap, int n, int p) {        int left = 2 * p + 1;        int right = 2 * p + 2;        int curIndex = p;        if ((left < n) && (heap[left]->val < heap[curIndex]->val)) {            curIndex = left;        }        if ((right < n) && (heap[right]->val < heap[curIndex]->val)) {            curIndex = right;        }        if (curIndex != p) {            ListNode* temp = heap[p];            heap[p] = heap[curIndex];            heap[curIndex] = temp;            MinHeapify(heap, n, curIndex);        }    }    void BuildMinHeap(vector<ListNode*>& heap, int n) {        for (int i = (n - 1) / 2; i >= 0; --i) {            MinHeapify(heap, n, i);        }    }    vector<ListNode*> heap;};