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]

.
题意：求已排序数组出现某个数的范围。
思路：两次二分，找出左右边界

class Solution {
	public:
		vector<int> searchRange(int A[], int n, int target) {
			int left = 0, right = n;
			while (left < right) {
				int mid = left + right >> 1;
				if (A[mid] >= target) 
					right = mid;
				else left = mid + 1;
			}
			int ansLeft = left;

			left = 0, right = n;
			while (left < right) {
				int mid = left + right >> 1;
				if (A[mid] > target)
					right = mid;
				else left = mid + 1;
			}

			if (A[ansLeft] != target)
				return vector<int>{-1, -1};
			else return vector<int>{ansLeft, left-1};
		}
};