Merge k Sorted Lists

看了网上的讲解

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

一开始想一个一个去合并，复杂度是 O(nk), n 是所有元素的个数。 

使用归并排序的思想，可以降到 o(n*log(k)). 

代码参考了网上的，写的很好，得学会

merge 就相当于得到已经排序好了的， mergeTwoLists 是合并

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param a list of ListNode    # @return a ListNode    def mergeTwoLists(self, l1, l2):          tra,trb = l1,l2          trc = ListNode(1000)          prehead = trc          while True:              if tra==None:                  trc.next = trb                  return prehead.next              if trb==None:                  trc.next = tra                  return prehead.next                            if tra.val<=trb.val:                  trc.next = tra                  tra = tra.next              else:                  trc.next = trb                  trb = trb.next              trc = trc.next                def merge(self,lists,l,r):        if l<r:            m = (l+r)/2            return self.mergeTwoLists(self.merge(lists,l,m),self.merge(lists,m+1,r))        return lists[l]        def mergeKLists(self, lists):        if len(lists)==0:            return None                return self.merge(lists,0,len(lists)-1)

c++

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {          ListNode *tra, *trb, *prehead;          tra = l1;          trb = l2;          ListNode *trc = new ListNode(100);          prehead = trc;          while (true){              if (tra==NULL){                  trc->next = trb;                  return prehead->next;              }              if (trb==NULL){                  trc->next = tra;                  return prehead->next;              }                            if (tra->val<=trb->val){                  trc->next = tra;                  trc = tra;                  tra = tra->next;              }              else{                  trc->next = trb;                  trc = trb;                  trb = trb->next;              }          }      }        ListNode *merge(vector<ListNode *> &lists, int l, int r){        if (l<r){            int m = (l+r)/2;            return mergeTwoLists(merge(lists,l,m),merge(lists,m+1,r));        }        return lists[l];    }        ListNode *mergeKLists(vector<ListNode *> &lists) {        if (lists.size()==0){            return NULL;        }                return merge(lists,0,lists.size()-1);    }};