LeetCode:Search for a Range

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].

其实这道题很简单，就是二分查到一个位置，在前后查找与target这个Range。

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public class Solution {
     public int[] searchRange(int[] A, int target) {
        int[] index = {-1,-1};
        if(A == null|| A.length == 0)return index;
        int left = 0;
        int right = A.length - 1;
        int mid = (left + right)/2;
        while(left<=right)
        {
            if(A[mid] == target)
            {
                index = range(A,mid);
                return index;
            }
            else if(A[mid]<target)
            {
                left = mid + 1;
            }
            else
            {
                right = mid - 1;
            }
            mid = (left + right)/2;
        }
        return index;
    }
    public int[] range(int[] A, int index)
    {
        int[] result = {index,index};
        int left = index - 1;
        int right = index + 1;
        while(left>=0&&A[left] == A[index])
        {
            result[0] = left;
            left = left - 1;
        }
        while(right<A.length&&A[right] == A[index])
        {
            result[1] = right;
            right = right + 1;
        }
        return result;
    }
}