LeetCode 23. Merge k Sorted Lists (单链表 + 递归)

时间:2022-01-25 19:47:39

这道题我是用递归做的,注定了只打败了一半的人-  -  讲真,这道题和MergeSort根本没有任何差别啊

不过我的是尾递归,尾递归总是可以改成while循环的。改了的话应该会快很多吧。

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode * res = NULL;
        if(lists.size() == 0)
            return res;
        int len;
        len = lists.size() - 1;
        res = mergeLists(lists, 0, len);
        return res;
    }
    
    ListNode * mergeLists(vector<ListNode *> & lists, int begin, int end){
        if(begin > end)
            return NULL;
        if(begin == end)
            return lists[begin];

        ListNode *p1 =  mergeLists(lists, begin, (begin + end)/2 );  //递归
        ListNode *p2 = mergeLists(lists, (begin+ end)/2  + 1, end);  //递归
        return mergeTwoLists(p1,p2);
    }
    
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {  //套用了上一题的答案
        struct ListNode * res = NULL;
        struct ListNode * p1 = l1;  //p1  p2用来跟踪l1,l2; p3永远指向新的链表的最后一个元素
        struct ListNode * p2 = l2;
        struct ListNode * p3 = res;
        while(p1 != NULL && p2 != NULL){
            int x ;
            if( p1->val < p2->val ){
                x = p1->val;
                p1 = p1->next;
            }
            else{
                x = p2->val;
                p2 = p2->next;
            }
            struct ListNode * temp = new ListNode(x);   //新建一个节点
            
            if(p3 == NULL){  //没有头节点就是很烦啊,各种NULL的判断
                res = temp;
                p3 = res;
            }
            else{
                p3->next = temp;
                p3 = p3->next;
            }
        }
        //一个结束了另一个还没有结束 
        if(p1 == NULL && p2 != NULL){
            if(p3 == NULL)    //一个都没加进去- -
                res = p2;
            else
                p3->next = p2;
        }
        else if(p1 != NULL && p2 == NULL){
            if(p3 == NULL)   //同样,一个都没加进去
                res = p1;
            else
                p3->next = p1;
        }
        
        return res;
        
    }
    
};