题解 | 合并k个已排序的链表

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <list>
#include <vector>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param lists ListNode类vector
     * @return ListNode类
     */
    ListNode* MergeTwoList(ListNode* pHead1, ListNode* pHead2) {
        // 考虑特殊情况
        if (!pHead1 || !pHead2)
            return pHead1 == nullptr ? pHead2 : pHead1;
        ListNode* dummy = new ListNode(0);
        ListNode* cur = dummy;
        // 双指针
        ListNode* p1 = pHead1, * p2 = pHead2;
        while (p1 && p2) {
            if (p1->val <= p2->val) {
                cur->next = p1;
                p1 = p1->next;
            } else {
                cur->next = p2;
                p2 = p2->next;
            }
            cur = cur->next;
        }
        // 若有一边余下，则直接连至末尾
        if (p1) cur->next = p1;
        if (p2) cur->next = p2;
        // C++需手动删除虚表头
        ListNode* newHead = dummy->next;
        delete dummy;
        return newHead;
    }
    ListNode* Merge(vector<ListNode*>& lists, int left, int right) {
        if (left == right) return lists[left];
        if (left > right) return nullptr;
        int mid = (left + right) >> 1;
        return MergeTwoList(Merge(lists, left, mid), Merge(lists, mid + 1, right));
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        // 思路：分治。分解为多个子问题后，两两合并
        return Merge(lists, 0, lists.size() - 1);
    }
};