描述：
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
思路：
1.二分查找，找到value為target的一個元素的位置，找不到返回[-1,-1]
2.假設找到位置，且為index，分別向兩邊拓展即可。
代碼：
public int[] searchRange(int[] A, int target) {
int arr[]=new int[2];
Arrays.fill(arr, -1);
if(A==null||A.length==0)
return arr;
int len=A.length;
int start=0,end=len-1;
int mid=0;
int index=-1;
while(start<=end)
{
mid=(start+end)/2;
if(A[mid]==target)
{
index=mid;
break;
}else if(A[mid]=0&&A[temp]==target)
{
i++;
temp=index-i;
}
arr[0]=index-i+1;
i=1;
temp=index+i;
while(temp
