LeetCode Search for a Range

题目：

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].

class Solution {
public:
	vector<int> searchRange(int A[], int n, int target) {
		vector<int> ans(2, -1);
		ans[0] = findLeft(A, n, target);
		ans[1] = findRight(A, n, target);
		return ans;
	}
private:
	int findLeft(int A[], int n, int target) {
		int left = -1;
		int low = 0, high = n - 1;
		while (low <= high) {
			int mid = (low + high) / 2;
			if (A[mid] < target) {
				low = mid+1;
			}
			else {
				if (A[mid] == target)
					left = mid;
				high = mid-1;	
			}
		}
		return left;
	}
	int findRight(int A[], int n, int target) {
		int right = -1;
		int low = 0, high = n - 1;
		while (low <= high) {
			int mid = (low + high) / 2;
			if (A[mid] <= target) {
				if (A[mid] == target)
					right = mid;
				low = mid + 1;

			}
			else
				high = mid-1;
		}
		return right;
	}
};