题目描述
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/merge-k-sorted-lists
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
方法1:顺序合并
思路
跟 21.合并两个有序链表 的基本思路是一样的,只不过每轮操作需要比较 k 个链表节点的值。
复杂度分析
- 时间复杂度:O(k\*n),k 是链表个数,n 是合并后链表的长度。
- 空间复杂度:O(1)。
代码(JavaScript/C++)
JavaScript Code
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode[]} lists
* @return {ListNode}
*/
var mergeKLists = function (lists) {
if (!lists || !lists.length) return null;
let dummy = new ListNode();
let tail = dummy;
while (!isEmpty(lists)) {
const min = getMin(lists);
tail.next = new ListNode(min.val);
tail = tail.next;
}
return dummy.next;
// **********************************************
function getMin(lists) {
let minIndex = -1,
minNode = new ListNode(Infinity);
for (let i = 0; i < lists.length; i++) {
const node = lists[i];
if (node && node.val < minNode.val) {
minNode = node;
minIndex = i;
}
}
lists[minIndex] = minNode.next;
return minNode;
}
function isEmpty(lists) {
return lists.every(n => n === null);
}
};
C++ code
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
private:
bool isEmpty_(vector<ListNode*>& lists) {
for (int i = 0; i < lists.size(); i++) {
if (lists[i]) return false;
}
return true;
}
int getMinVal_(vector<ListNode*>& lists) {
int ans = INT_MAX;
int idx = -1;
for (int i = 0; i < lists.size(); i++) {
if (lists[i] && lists[i]->val < ans) {
ans = lists[i]->val;
idx = i;
}
}
lists[idx] = lists[idx]->next;
return ans;
}
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
ListNode* dummy = new ListNode();
ListNode* tail = dummy;
while (!isEmpty_(lists)) {
int min_val = getMinVal_(lists);
tail->next = new ListNode(min_val);
tail = tail->next;
}
return dummy->next;
}
};
方法2:顺序合并+堆
思路
跟 方法1 思路一致,不过用了堆来寻找 k 个链表节点中的最小值。
复杂度分析
- 时间复杂度:O(nlogk),k 是链表个数,n 是合并后链表的长度。
- 空间复杂度:O(k),k 是链表个数,堆的空间大小。
代码
JavaScript Code
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode[]} lists
* @return {ListNode}
*/
var mergeKLists = function (lists) {
if (!lists || !lists.length) return null;
let dummy = new ListNode();
let tail = dummy;
const heap = new MinHeap(lists.filter(Boolean), function comparator(inserted, compared) {
return inserted.val > compared.val;
});
while (heap.size() > 0) {
const min = heap.pop();
tail.next = new ListNode(min.val);
tail = tail.next;
min.next && heap.insert(min.next)
}
return dummy.next;
};
// **************************************************
class Heap {
constructor(list = [], comparator) {
this.list = list;
this.comparator = comparator;
this.init();
}
init() {
const size = this.size();
for (let i = Math.floor(size / 2) - 1; i >= 0; i--) {
this.heapify(this.list, size, i);
}
}
insert(n) {
this.list.push(n);
const size = this.size();
for (let i = Math.floor(size / 2) - 1; i >= 0; i--) {
this.heapify(this.list, size, i);
}
}
peek() {
return this.list[0];
}
pop() {
const last = this.list.pop();
if (this.size() === 0) return last;
const returnItem = this.list[0];
this.list[0] = last;
this.heapify(this.list, this.size(), 0);
return returnItem;
}
size() {
return this.list.length;
}
}
class MinHeap extends Heap {
constructor(list, comparator) {
if (typeof comparator != 'function') {
comparator = function comparator(inserted, compared) {
return inserted > compared;
};
}
super(list, comparator);
}
heapify(arr, size, i) {
let smallest = i;
const left = Math.floor(i * 2 + 1);
const right = Math.floor(i * 2 + 2);
if (left < size && this.comparator(arr[smallest], arr[left]))
smallest = left;
if (right < size && this.comparator(arr[smallest], arr[right]))
smallest = right;
if (smallest !== i) {
[arr[smallest], arr[i]] = [arr[i], arr[smallest]];
this.heapify(arr, size, smallest);
}
}
}
方法3:分治
思路
- 将 k 个链表平均分成两份,分别进行合并后,再将两个结果进行合并。
- 最后一步就是简单的 合并两个有序链表。
- 而中间的步骤也很简单,只需要将 k/2 个链表再细分,再细分,细分到一次只需要处理两个链表就好了。
复杂度分析
- 时间复杂度:O(nlogk),k 是链表个数,n 是合并后链表的长度。
- 空间复杂度:O(k),k 是链表个数,堆的空间大小。
代码(JavaScript/C++)
JavaScript Code
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode[]} lists
* @return {ListNode}
*/
var mergeKLists = function(lists, start = 0, end = lists.length - 1) {
if (start > end) return null;
if (start === end) return lists[start];
const mid = ((end - start) >> 1) + start;
return mergeTwoLists(mergeKLists(lists, start, mid), mergeKLists(lists, mid + 1, end));
};
function mergeTwoLists(l1, l2) {
if (!l1) return l2;
if (!l2) return l1;
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
C++ Code
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
return mergeKLists(lists, 0, lists.size() - 1);
}
ListNode* mergeKLists(vector<ListNode*>& lists, int start, int end) {
if (start > end) return nullptr;
if (start == end) return lists[start];
int m = (end - start) / 2 + start;
return merge2Lists(mergeKLists(lists, start, m), mergeKLists(lists, m + 1, end));
}
ListNode* merge2Lists(ListNode* l1, ListNode* l2) {
if (!l1) return l2;
if (!l2) return l1;
if (l1->val < l2->val) {
l1->next = merge2Lists(l1->next, l2);
return l1;
} else {
l2->next = merge2Lists(l1, l2->next);
return l2;
}
}
};