[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();
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)>>;
ListNode *lleft = mergesort(lists,left,middle);
ListNode *lright = mergesort(lists,middle+,right);
return merge(lleft,lright);
} }
ListNode *mergeKLists(vector<ListNode *> &lists) {
int left =,right =lists.size()-;
int middle= (left+right)>>;
ListNode *lleft = mergesort(lists,left,middle);
ListNode *lright = mergesort(lists,middle+,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
转载请注明出处： http://www.cnblogs.com/double-win/谢谢！
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