leetcode_Search for a Range

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为target的一个元素的位置，找不到返回[-1,-1]

2.假设找到位置，且为index，分别向两边拓展即可。

代码：

 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]<target)        start=mid+1;        else         end=mid-1;        }        if(index==-1)        return arr;        int i=1,temp=index-i;        while(temp>=0&&A[temp]==target)        {        i++;        temp=index-i;        }        arr[0]=index-i+1;        i=1;        temp=index+i;        while(temp<len&&A[temp]==target)        {        i++;        temp=index+i;        }        arr[1]=index+i-1;        return arr;    }