Merge k Sorted Lists

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

 

思路：要快速找到k个元素中的最小值，使用堆这种数据结构。找到最小值后，还需要知道该最小值出自哪个链表，因此，堆中还需记录每个节点来自哪个链表。调整堆的时间复杂度为O(logk)，所以总体时间复杂度为O(nlogk)。代码如下：

 

struct ListNode {int val;struct ListNode *next;};typedef struct{    int val;    int index;}HeapNode;void swap(HeapNode *a, HeapNode *b){    HeapNode temp;    temp = *b;    *b = *a;    *a = temp;}void minheapify(HeapNode *heap, int heapsize, int index){    int left, right;    int key;        int smallest;    while(index < heapsize)    {        left = 2*index+1;        right = 2*index+2;        smallest = index;        if(left < heapsize && heap[index].val > heap[left].val)        {            smallest = left;        }        if(right < heapsize && heap[smallest].val > heap[right].val)        {            smallest = right;        }                if(smallest == index)   return;        swap(heap+smallest, heap+index);        index = smallest;    }}void buildminheap(HeapNode *heap, int heapsize){    int start = (heapsize-2)/2;        while(start >= 0)    {        minheapify(heap, heapsize, start);        start--;    }}struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {    int i ,j;    int heapsize = listsSize;    int minindex = -1;    struct ListNode *head = NULL;    struct ListNode *node = NULL;        if(listsSize == 0)  return head;    HeapNode *heap = calloc(heapsize, sizeof(HeapNode));    for(i = 0, j = 0; i < heapsize; i++)    {        if(lists[i] == NULL)        {            continue;        }        heap[j].val = lists[i]->val;        heap[j].index = i;        j++;    }    heapsize = j;    if(heapsize == 0)   return head;        buildminheap(heap, heapsize);    minindex = heap[0].index;    head = lists[minindex];    node = head;    lists[minindex] = lists[minindex]->next;    while(1)    {        if(lists[minindex] == NULL)        {            heapsize -= 1;            if(heapsize == 0)   return head;            swap(heap, heap+heapsize);            minheapify(heap, heapsize, 0);        }        else        {            heap[0].val = lists[minindex]->val;            minheapify(heap, heapsize, 0);        }        minindex = heap[0].index;        node->next = lists[minindex];        node = node->next;        lists[minindex] = lists[minindex]->next;    }}