Leetcode 题解系列（四）

23. Merge k Sorted Lists

题目要求

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

题目分析

1. 暴力解法

每一次遍历K个列表，取得最大的元素并进行归并。

ListNode* mergeKLists(vector<ListNode*>& lists) {  bool completed = false;  ListNode *ret = nullptr, *cur = nullptr;  while (!completed) {    int index, max = 0;    completed = true;    for (int i = 0; i < lists.size(); ++i) {      if (lists[i] != nullptr) {        completed = false;      } else {        continue;      }      if (lists[i]->val > max) {        max = lists[i]->val;        index = i;      }    }    if (cur == nullptr) {      ret = cur = lists[index];    } else {      cur->next = lists[index];      cur = cur->next;    }    lists[index] = lists[index]->next;  }  return ret;}

2. 使用堆进行加速

上述的方法需要进行N次的遍历，其中N为元素总数，故复杂度为O(N2)。使用堆进行维护，可以使得获取最大值的复杂度为O(1) + O(log(K))（取值与维护）。那么总的复杂度变为O(Nlog(K))

/** * Definition for singly-linked list. * struct ListNode { *   int val; *   ListNode *next; *   ListNode(int x) : val(x), next(NULL) {} * }; */#include <queue>struct ListCompare {  bool operator() (const ListNode* lhs, const ListNode* rhs) {    return lhs->val > rhs->val;  }};using pq = std::priority_queue<ListNode*, vector<ListNode*>, ListCompare>;class Solution {public:  ListNode* mergeKLists(vector<ListNode*>& lists) {    pq queue;    for (auto node : lists) {      if (node != nullptr) {        queue.push(node);      }    }    ListNode* ret = nullptr, *cur = nullptr;    while (!queue.empty()) {      auto one = queue.top();      queue.pop();      if (cur == nullptr) {        ret = cur = one;      } else {        cur->next = one;        cur = cur->next;      }      one = one->next;      if (one != nullptr) queue.push(one);    }    return ret;  }};

3. 使用分治的方法进行归并

对K个待归并的子链表进行分治归并，可以降低复杂度。