给你一个链表数组,每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中,返回合并后的链表。
示例 1:
输入:lists = [[1,4,5],[1,3,4],[2,6]]
输出:[1,1,2,3,4,4,5,6]
解释:链表数组如下:
[
1->4->5,
1->3->4,
2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2:
输入:lists = []
输出:[]
示例 3:
输入:lists = [[]]
输出:[]
提示:
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i]
按 升序 排列lists[i].length
的总和不超过10^4
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
auto cmp = [](ListNode* a, ListNode* b){return a->val > b->val;};
priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> minHeap(cmp);
for(int i = 0; i < lists.size(); ++i){
while(lists[i]){
minHeap.push(lists[i]);
lists[i] = lists[i]->next;
}
}
if(minHeap.empty()) return nullptr;
ListNode* res = minHeap.top();
ListNode* cur = res;
minHeap.pop();
while(!minHeap.empty()){
cur->next = minHeap.top();
minHeap.pop();
cur = cur->next;
}
cur->next = nullptr;
return res;
}
};
标签:ListNode,cur,23,lists,链表,升序,minHeap
From: https://blog.51cto.com/u_15862486/7529757