LeetCode-23 合并 K 个升序链表

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

solution

采用分治法，

从pre_vec合并两个相邻的链表，并将结果放入vec中
pre_vec=vec,vec.clear()
循环上述操作，直至pre_vec的长度为1
输出pre_vec[0]

#include <vector>using namespace std;//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) {}
};//leetcode submit region begin(Prohibit modification and deletion)
/*** 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:ListNode *mergeLists(ListNode *A, ListNode *B) {ListNode *L = new ListNode(), *P_A = A, *P_B = B, *cur = L;while (P_A != nullptr && P_B != nullptr) {if (P_A->val < P_B->val) {ListNode *P = new ListNode(P_A->val);cur->next = P;cur = cur->next;P_A = P_A->next;} else {ListNode *P = new ListNode(P_B->val);cur->next = P;cur = cur->next;P_B = P_B->next;}}if (P_A!= nullptr) {cur->next = P_A;}if (P_B!= nullptr) {cur->next = P_B;}return L->next;}public:ListNode *mergeKLists(vector<ListNode *> &lists) {vector<ListNode *> pre_vec = lists;vector<ListNode *> vec;if (lists.size()==0){return nullptr;}else if (lists.size()==1){return lists[0];}do {for (int i = 0; i < pre_vec.size(); i += 2) {if (i + 1 < pre_vec.size()) {vec.push_back(mergeLists(pre_vec[i], pre_vec[i + 1]));} else {vec.push_back(pre_vec[pre_vec.size() - 1]);}}pre_vec = vec;vec.clear();} while (pre_vec.size() != 1);return pre_vec[0];}
};
//leetcode submit region end(Prohibit modification and deletion)ListNode *createList(int a[], int n) {ListNode *A = new ListNode(), *cur = A;for (int i = 0; i < n; ++i) {ListNode *p = new ListNode(a[i]);p->next = NULL;cur->next = p;cur = cur->next;}return A->next;
}int main() {int a[3] = {1, 4, 5}, b[3] = {1, 3, 4}, c[2] = {2, 3}, na = 3, nb = 3, nc = 2;vector<ListNode *> A = {createList(a, na), createList(b, nb), createList(c, nc)};Solution solution;solution.mergeKLists(A);}