Merge k Sorted Lists leetcode java

题目：

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

题解：

Merge k sorted linked list就是merge 2 sorted linked list的变形题。

而且我们很自然的就想到了经典的Merge Sort，只不过那个是对数组进行sort。而不同的地方，仅仅是Merge两个list的操作不同。

这里来复习一下Merge Sort(对于数组操作），参考Wikipedia：

归并操作(merge)，也叫归并算法，指的是将两个已经排序的序列合并成一个序列的操作。归并排序算法依赖归并操作。

归并操作的过程如下：

申请空间，使其大小为两个已经排序序列之和，该空间用来存放合并后的序列
设定两个指针，最初位置分别为两个已经排序序列的起始位置
比较两个指针所指向的元素，选择相对小的元素放入到合并空间，并移动指针到下一位置
重复步骤3直到某一指针到达序列尾
将另一序列剩下的所有元素直接复制到合并序列尾

最差時間複雜度
最優時間複雜度
平均時間複雜度
最差空間複雜度

Merge Sort 对数组操作的Java代码为：

1 public int[] mergeSort(int[] arr){
2     if(arr.length<2||arr == null)
3         return arr;
4
5     MSort(arr,0,arr.length-1);
6 }
7
8 public int[] MSort(int[] arr, int low, int high){
9         if(low < high){
10                 int mid = (low+high)/2;
11                 int[] left = MSort(arr,low,mid);
12                 int[] right = MSort(arr,mid+1,high);
13                 return mergeTwoList(left,right);
14           }
15 }
16
17
18 public int[] mergeTwoList(int[] A, int[] B) {
19     int[] C = new int[A.length + B.length];
20     int k = 0;
21     int i = 0;
22     int j = 0;
23     while(i < A.length && j < B.length) {
24         if (A[i] < B[j])
25             C[k++] = A[i++];
26         else
27             C[k++] = B[j++];
28     }
29     while (i < A.length)
30         C[k++] = A[i++];
31     while (j < B.length)
32         C[k++] = B[j++];
33     return C;
34 }

下面就是这道题的解法，跟上面的方法一样一样的，就是在mergeTwoList时候是对linkedlist做，套用Merge 2 sorted list解法即可，代码如下：

1     public ListNode mergeKLists(ArrayList<ListNode> lists) {
2         if(lists==null || lists.size()==0)
3             return null;
4         return MSort(lists,0,lists.size()-1);
5     }
6
7     public ListNode MSort(ArrayList<ListNode> lists, int low, int high){
8         if(low < high){
9             int mid = (low+high)/2;
10             ListNode leftlist = MSort(lists,low,mid);
11             ListNode rightlist = MSort(lists,mid+1,high);
12             return mergeTwoLists(leftlist,rightlist);
13         }
14         return lists.get(low);
15     }
16
17
18     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
19             if(l1==null)
20                 return l2;
21             if(l2==null)
22                 return l1;
23
24             ListNode l3;
25             if(l1.val<l2.val){
26                 l3 = l1;
27                 l1 = l1.next;
28             }else{
29                 l3 = l2;
30                 l2 = l2.next;
31             }
32
33             ListNode fakehead = new ListNode(-1);
34             fakehead.next = l3;
35             while(l1!=null&&l2!=null){
36                 if(l1.val<l2.val){
37                     l3.next = l1;
38                     l3 = l3.next;
39                     l1 = l1.next;
40                 }else{
41                     l3.next = l2;
42                     l3 = l3.next;
43                     l2 = l2.next;
44                 }
45             }
46
47             if(l1!=null)
48                 l3.next = l1;
49             if(l2!=null)
50                 l3.next = l2;
51             return fakehead.next;
52         }

更多Mergesort的讲法请参考：http://www.cs.princeton.edu/courses/archive/spr07/cos226/lectures/04MergeQuick.pdf

讲的挺好的

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