Algorithm

[LeetCode 34] Search for a Range [Java]

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2018-01-25

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1. Description

Given an array of integers sorted in ascending order, 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].

2. Example

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

3. Code


public class LeetCode0034 {
    public int[] searchRange(int[] nums, int target) {
        int[] result = new int[] { -1, -1 };
        if (nums == null || nums.length == 0) {
            return result;
        }

        int left = 0, right = nums.length - 1, mid = -1;

        while (left <= right) {
            mid = left + ((right - left) >> 1);
            if (nums[mid] == target) {
                break;
            }
            if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        if (left > right) {
            return result;
        }

        for (left = mid; left >= 0 && nums[left] == target; left--)
            ;
        for (right = mid; right < nums.length && nums[right] == target; right++)
            ;

        result[0] = left < 0 ? 0 : nums[left] == target ? left : left + 1;
        result[1] = right == nums.length ? nums.length - 1 : nums[right] == target ? right : right - 1;
        return result;
    }

    public static void main(String[] args) {
        LeetCode0034 leetcode = new LeetCode0034();
        int[] result = leetcode.searchRange(new int[] { 5, 7, 7, 8, 8, 10 }, 9);
        System.out.println(result[0] + " , " + result[1]);
    }
}

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