牛客BM5. 合并k个已排序的链表

Description:

题目大意：合并 K 个已经排序的链表

解题思路:

算法标签：分治

1. 将 k 个链表分成两两一组
2. 然后两个链表合并
3. 最终递归合并成为一个链表

代码:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    // 两个链表合并
    ListNode* mergeTowLists(ListNode* list1 , ListNode* list2) {
        if(list1 == NULL)
            return list2;
        if(list2 == NULL) 
            return list1;
        
        if(list1 -> val <= list2 -> val) {
            list1 -> next = mergeTowLists(list1 -> next, list2);
            return list1;
        }
        else {
            list2 -> next = mergeTowLists(list1 , list2 -> next);
            return list2;
        }
    }
    // 对每一个小组链表进行合并
    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 mergeTowLists(merge(lists , left , mid) , merge(lists , mid+1 , right));
    }
    ListNode* mergeKLists(vector<ListNode *> &lists) {
        int left = 0,right = lists.size() - 1;
        return merge(lists , left , right);
    }
};