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,

[5, 7, 7, 8, 8, 10] and target value 8,return [3, 4].

public int[] searchRange(int[] nums, int target) {        int low = 0, high = nums.length - 1, temp = -1;        int result[] = new int[2];        while (low <= high) {            int mid = (low + high) / 2;            if (nums[mid] == target) {                temp = mid;                break;            }            if (nums[mid] > target)                high = mid - 1;            if (nums[mid] < target)                low = mid + 1;        }        if (temp == -1) {            Arrays.fill(result, -1);            return result;        }        int tmp = temp;        while (tmp > 0 && target == nums[tmp - 1])            tmp--;        while (temp < nums.length-1 && target == nums[temp + 1])            temp++;        result[0] = tmp;        result[1] = temp;        return result;    }

上discuss看了简单一些的方法吧，分享下。

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