合并K个升序链表

思路

• 一分为二，递归求解
• 最后合并
• 子问题 ： 两个升序链表的归并

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        if(lists.size() == 0) return nullptr;
        if(lists.size() == 1) return lists[0];
        // 一分为2
        int mid = lists.size() / 2;  
        vector<ListNode*> left = vector<ListNode*> (lists.begin(), lists.begin() + mid);
        vector<ListNode*> right = vector<ListNode*> (lists.begin() + mid, lists.end());  
        // 递归处理
        ListNode *l1 = mergeKLists(left), *l2 = mergeKLists(right);
        // 最后归并
        return merge2lists(l1, l2);
    }
    
     ListNode *merge2lists(ListNode *l1, ListNode* l2)
    {
        ListNode *dummy = new ListNode(-1);
        ListNode *cur = dummy;
        
        while(l1 && l2)
        {
            if(l1 -> val <= l2 -> val)
            {
                cur = cur -> next = l1;
                l1 = l1 -> next;
            }
            else
            {
                cur  = cur -> next = l2;
                l2  = l2 -> next;
            }
        }
        
        cur -> next = l1 ? l1 : l2;
        
        return dummy -> next;
    }
};