【leetcode】Merge k Sorted Lists

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

O(N * logK)

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode **g_lists;int g_num;#define max_num 10000struct ListNode *g_heap[max_num];int heap_num;void rebuildHeap(int i) {    int left = i * 2 + 1;    int right = i * 2 + 2;    if (left >= heap_num) {        return;    }    int min = i;    if (g_heap[i]->val > g_heap[left]->val) {        min = left;    }    if (right < heap_num && g_heap[min]->val > g_heap[right]->val) {        min = right;    }    if (min == i) {        return;    }    struct ListNode *tmp = g_heap[i];    g_heap[i] = g_heap[min];    g_heap[min] = tmp;    rebuildHeap(min);}void makeHeap() {    for (int i = heap_num / 2; i >= 0; --i) {        rebuildHeap(i);    }}struct ListNode *pickOne() {    if (heap_num == 0) {        return NULL;    }    struct ListNode *min = g_heap[0];    if (min->next == NULL) {        heap_num--;        g_heap[0] = g_heap[heap_num];    } else {        g_heap[0] = g_heap[0]->next;    }    rebuildHeap(0);    min->next = NULL;    return min;}struct ListNode *mergeKLists(struct ListNode *lists[], int k) {    if (k == 0) {        return NULL;    }    if (k == 1) {        return lists[0];    }    g_num = k;    g_lists = lists;    heap_num = 0;    for (int i = 0; i < g_num; ++i) {        if (g_lists[i] != NULL) {            g_heap[heap_num++] = g_lists[i];        }    }    makeHeap();        struct ListNode *head = NULL;    struct ListNode *cur = NULL;    struct ListNode *p = NULL;        while ((p = pickOne()) != NULL) {        if (head == NULL) {            head = p;        } else {            cur->next = p;        }        cur = p;    }        return head;}