23. Merge k Sorted Lists

1.問題

• Merge多個sorted list並回傳完整的sorted list.

• 分析並解釋他的Complexity

2.想法

• Merge sort, 省略掉split的部分, 將每個sorted list當成最小的單位

3.程式碼

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* merge(ListNode* l1, ListNode *l2){
        if(!l1){
            return l2;
        }
        
        if(!l2){
            return l1;
        }
        
        ListNode *res = nullptr;
        ListNode **ptr;
        
        if(l1->val < l2->val){
            res = l1;
            l1 = l1->next;
        }else{
            res = l2;
            l2 = l2->next;
        }
        
        ptr = &res;
        
        //Merge 2 list
        while(l1 && l2){
            if(l1->val < l2->val){
                (*ptr)->next = l1;
                l1 = l1->next;
            }else{
                (*ptr)->next = l2;
                l2 = l2->next;
            }
            ptr = &((*ptr)->next);
        }
        
        if(l1){
            (*ptr)->next = l1;
        }else if(l2){
            (*ptr)->next = l2;
        }
                    
        return res;
    }
    
    ListNode *mergeK(vector<ListNode*>::iterator b, vector<ListNode*>::iterator e){
        if(b == e){
            return nullptr;
        }
        
        if(next(b) == e){
            return *b;
        }
        
        int mid = distance(b,e)/2;
        
        ListNode *left= mergeK(b,b + mid);
        ListNode *right = mergeK(b + mid,e);
        return merge(left,right);
    }
    
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        return mergeK(lists.begin(),lists.end());
    }
};

4.Performance