Java for LeetCode 034 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].

解题思路：

看到O(log n) 几乎可以肯定是二分查找的思路，题目不是特别难的那种，仔细想想就想出来了，JAVA实现如下：

	static public int[] searchRange(int[] nums, int target) {
		int[] result = new int[2];
		result[0] = result[1] = -1;
		int left = 0, right = nums.length - 1;
		while (left <= right) {
			if (nums[(left + right) / 2] > target)
				right = (left + right) / 2 - 1;
			else if (nums[(left + right) / 2] < target)
				left = (left + right) / 2 + 1;
			else {
				result[0] = result[1] = (left + right) / 2;
				while (target != nums[left]) {
					if (target > nums[(result[0] + left) / 2])
						left = (result[0] + left) / 2 + 1;
					else {
						result[0] = (result[0] + left) / 2;
						left++;
					}
				}
				result[0] = left;
				while (target != nums[right]) {
					if (target < nums[(result[1] + right) / 2])
						right = (result[1] + right) / 2 - 1;
					else {
						result[1] = (result[1] + right) / 2;
						right--;
					}
				}
				result[1] = right;
				break;
			}
		}
		return result;
	}