Merge Sorted Array leetcode java（回顾MergeTwoArray和MergeTwoLinkedList）

题目：

Given two sorted integer arrays A and B, merge B into A as one sorted array.

Note:
You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m and n respectively.

题解：

这道题是说让B merge到 A 里面。

先复习下原本我们在MergeSort里面怎么利用一个新建的数量来merge two array：

代码如下：

 1 public int[] mergeTwoList(int[] A, int[] B) {
 2     int[] C = new int[A.length + B.length];
 3     int k = 0;
 4     int i = 0;
 5     int j = 0;
 6     while(i < A.length && j < B.length) {
 7         if (A[i] < B[j])
 8             C[k++] = A[i++];
 9         else
10             C[k++] = B[j++];
11     }
12     while (i < A.length) 
13         C[k++] = A[i++];
14     while (j < B.length) 
15         C[k++] = B[j++];
16     return C;
17 }

然后我们再顺便复习下，怎么merge two linked list，代码如下：

 1     public ListNode mergeTwoLists(ListNode leftlist, ListNode rightlist){
 2         if(rightlist == null)
 3             return leftlist;
 4         if(leftlist == null)
 5             return rightlist;
 6         
 7         ListNode fakehead = new ListNode(-1);
 8         ListNode ptr = fakehead;
 9         while(rightlist!=null&&leftlist!=null){
10             if(rightlist.val<leftlist.val){
11                 ptr.next = rightlist;
12                 ptr = ptr.next;
13                 rightlist = rightlist.next;
14             }else{
15                 ptr.next = leftlist;
16                 ptr = ptr.next;
17                 leftlist = leftlist.next;
18             }
19         }
20         
21         if(rightlist!=null)
22             ptr.next = rightlist;
23         if(leftlist!=null)
24             ptr.next = leftlist;
25         
26         return fakehead.next;
27     }

可以看出merge的思路都是在从头比较两个list的value，用两个指针分别指向当前要比较的node上面。而且最后都会处理下剩下的元素。

而这道题是不能借助一个新的array的，那么我们就不好从前往后比了（不好插入位置）。方便的方法是从后往前比，然后最后处理剩下的元素。

代码如下：

     public void merge(int A[], int m, int B[], int n) {
        while(m > 0 && n > 0){
            if(A[m-1] > B[n-1]){
                A[m+n-1] = A[m-1];
                m--;
            }else{
                A[m+n-1] = B[n-1];
                n--;
            }
        }
 
        while(n > 0){
            A[m+n-1] = B[n-1];
            n--;
        }
    }