[LeetCode]--34. 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].

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 };
    }