leetcode 日经贴，Cpp code -Merge k Sorted Lists

最新推荐文章于 2021-02-25 11:39:59 发布

原创最新推荐文章于 2021-02-25 11:39:59 发布 · 315 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 3 个专栏收录

183 篇文章

订阅专栏

leetcode

173 篇文章

订阅专栏

Cpp

152 篇文章

订阅专栏

本文介绍了一种使用优先级队列合并多个已排序链表的方法。通过定义一个自定义节点结构体并重载比较运算符，可以高效地将来自不同链表的节点按顺序合并成一个单一的排序链表。

Merge k Sorted Lists

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

struct node {
    ListNode *lnp;
    node() {
        lnp = NULL;
    }
    node (ListNode *_lnp): lnp(_lnp){}
    bool operator < (const node &other) const {
        return lnp->val > other.lnp->val;
    }
};

class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        priority_queue<node> pq;
        ListNode *head = NULL, *current = NULL;
        int n = lists.size();
        for (int i = 0; i < n; ++i) {
            if (lists[i]) {
                pq.push(node(lists[i]));
            }
        }
        while (!pq.empty()) {
            ListNode *p = pq.top().lnp;
            if (!head) {
                head = current = p;
            } else {
                current->next = p;
                current = p;
            }
            pq.pop();
            if (p->next) {
                pq.push(node(p->next));
            }
        }
        return head;
    }
};

Using STL

/**
 * 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) {
        ListNode *head = NULL, *cur = NULL;
        priority_queue<pair<int, ListNode*>, vector<pair<int, ListNode*> >, greater<pair<int, ListNode*> > > pq;
        for (int i = 0; i < lists.size(); ++i) {
            if (lists[i]) {
                pq.push(make_pair(lists[i]->val, lists[i]));
            }
        }
        while (!pq.empty()) {
            ListNode *next = pq.top().second;
            pq.pop();
            if (!head) {
                head = cur = next;
            } else {
                cur->next = next;
                cur = next;
            }
            if (next->next) {
                pq.push(make_pair(next->next->val, next->next));
            }
        }
        return head;
    }
};