LeetCode-23 合并 K 个升序链表

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，

1. 从pre_vec合并两个相邻的链表，并将结果放入vec中
2. pre_vec=vec,vec.clear()
3. 循环上述操作，直至pre_vec的长度为1
4. 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);

}