21.23. Merge Two Sorted Lists /

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

思路：选择两个list中第一个元素最小的为基础，然后将另外一个list的元素逐渐和第一个list中的元素相比，并将其插入到第一个list中，最终返回第一个list的头即可。

时间复杂度：O(M+N)

空间复杂度：O(M+N)

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {              if(l1==null)            return l2;        if(l2==null)            return l1;        if(l1.val>l2.val)            return mergeTwoLists(l2,l1);                ListNode head=l1;        ListNode temp;        while(l2!=null){            while(l1.next!=null&&l2.val>l1.next.val)                l1=l1.next;            if(l1.next==null){                l1.next=l2;                return head;            }else{                temp=l2;                l2=l2.next;                temp.next=l1.next;                l1.next=temp;                l1=l1.next;                }                    }                return head;        }

递归算法：

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {              if(l1==null)            return l2;        if(l2==null)            return l1;                if(l1.val<l2.val){            l1.next=mergeTwoLists(l1.next,l2);            return l1;        }else{            l2.next=mergeTwoLists(l1,l2.next);            return l2;        }        }

23. Merge k Sorted Lists

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

思路: 两两合并,每次将两个listnode合并成一个. 利用到上述实现2个有序list合并的代码.

时间复杂度:O(N*logK) N为所有lists所包含的节点个数

 public ListNode mergeKLists(ListNode[] lists) {        int len=lists.length;        if(lists==null||len==0)            return null;               return mergeKList(lists,0,len-1);    }        public ListNode mergeKList(ListNode[] lists,int left, int right) {        int len=right-left;                if(len==0)            return lists[left];        else if(len==1){            return mergeTwoLists(lists[left],lists[right]);        }else{            return mergeTwoLists(mergeKList(lists,left,left+len/2),mergeKList(lists,len/2+left+1,right));        }    }                    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {                          if(l1==null)                  return l2;              if(l2==null)                  return l1;                            if(l1.val<l2.val){                  l1.next=mergeTwoLists(l1.next,l2);                  return l1;              }else{                  l2.next=mergeTwoLists(l1,l2.next);                  return l2;              }                    }