Search for a Range.txt

problem：
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].
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solution：
//二分查找的进阶版，当找左界时，A[mid] == target下，仍对左边继续二分；同理当找右界时，相等仍对右边二分
int findIndex(int A[], int l, int r, int target, bool isLeft)
{
  if(l > r)
		return -1;
	int mid = (l + r) / 2;
	if(A[mid] == target)
	{
		int pos = isLeft ? findIndex(A, l, mid - 1, target, isLeft) : findIndex(A, mid + 1, r, target, isLeft);
		return pos == -1 ? mid : pos;
	}
	else if(A[mid] < target)
		return findIndex(A, mid + 1, r, target, isLeft);
	else
		return findIndex(A, l, mid - 1, target, isLeft);
}

vector<int> searchRange(int A[], int n, int target)
{
  vector<int> ret(2, -1);
	if(n <= 0)
		return ret;
	if(target <A[0] || target > A[n-1])
		return ret;
	bool isLeft = true;
	int leftIndex = findIndex(A, 0, n-1, target, isLeft);
	if(leftIndex == -1)
		return ret;
	int rightIndex = findIndex(A, 0, n -1, target, !isLeft);
	ret[0] = leftIndex;
	ret[1] = rightIndex;
	return ret;
}