[LeetCode 题解]: Merge k Sorted Lists

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

题意：对k个有序的链表进行归并排序。并分析其复杂度。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* merge(ListNode* left, ListNode* right){   //归并操作
        if(left==NULL) return right;
        if(right==NULL) return left;
        ListNode *ans =new ListNode(0);
        if(left->val <= right->val){
            ans=left;
            left = left->next;    
        }else{
            ans=right;
            right=right->next;
        }
        ans->next = merge(left,right);
        return ans;
    }
    ListNode* mergesort(vector<ListNode *> &lists,int left,int right){  //devide && conquer
        if(left>right){
            return NULL;
        }if(left==right){
            return lists[left];
        }else{
            int middle = (left+right)>>1;
            ListNode *lleft = mergesort(lists,left,middle);
            ListNode *lright = mergesort(lists,middle+1,right);
            return merge(lleft,lright);
        }
        
    }
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        int left =0,right =lists.size()-1;
        int middle= (left+right)>>1;
        ListNode *lleft = mergesort(lists,left,middle);
        ListNode *lright = mergesort(lists,middle+1,right);
        return merge(lleft,lright);
    }
};

时间复杂度： O(n*logn) 

空间复杂度： O(1)

至于具体的量化分析呢，呵呵。。。数学推导了

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