[LeetCode] 23. Merge k Sorted Lists

思路一:
二分法merge, 我用的方法是每次首尾的链表merge, 然后结果存在原首链表的位置上, 这里要用到前面的merge two sorted lists, 忘记的朋友可以复习一下. 算法总时间复杂度O(nklogk).

ListNode* merge2Lists(ListNode* l1, ListNode* l2) {    ListNode* dummy = new ListNode(0);    ListNode* dummySaver = dummy;    while (l1 && l2) {        if (l1->val < l2->val)            dummy->next = l1, l1 = l1->next;        else             dummy->next = l2, l2 = l2->next;        dummy = dummy->next;    }    if (l1) dummy->next = l1;    if (l2) dummy->next = l2;    ListNode* head = dummySaver->next;    delete dummySaver;    return head;}ListNode* mergeKLists(vector<ListNode*>& lists) {    int size = lists.size();    if (! size) return NULL;    while (size > 1) {        int mid = size >> 1;        for (int i = 0; i < mid; i++)             // 收尾链表merge            lists[i] = merge2Lists(lists[i],                         lists[size - i - 1]);        // 如果是有奇数个链表, merge完一轮还剩n / 2 + 1个        if (size & 1) size = (size >> 1) + 1;        // 如果是有偶数个链表, merge完一轮还剩n / 2个        else size >>= 1;    }    return lists[0];}

思路二:
维护一个最小堆(优先队列). 先把所有链表的首元素push进队列, 然后每pop一个节点就把这个节点的后节点再push进队列, 队列最多会保持k个元素, 如果哪个链表遍历完了, 那么队列的元素就会减少, 一直循环, 直到队列没有元素为止. 每次队列pop出的元素就放到我们的结果链表尾即可. 时间复杂度O(nklogk).

class NodeCmp {public:    bool operator()(ListNode* l1, ListNode* l2) {        return l1->val > l2->val;    }};ListNode* mergeKLists(vector<ListNode*>& lists) {    priority_queue<ListNode*,                 vector<ListNode*>,                NodeCmp> heap;    for (int i = 0; i < lists.size(); i++)         if (lists[i]) heap.push(lists[i]);    ListNode dummy = ListNode(0);    ListNode* pDummy = &dummy;    ListNode* saveDummy = pDummy;    while (heap.size()) {        pDummy->next = heap.top();        pDummy = pDummy->next;        ListNode* tempNext = heap.top()->next;        heap.pop();        if (tempNext) heap.push(tempNext);     }    return saveDummy->next;}

这是到很有趣也经常会用到的问题, 两种方法都推荐学会并记住.