34 Search for a Range

34. Search for a Range

1. Question

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(logn).

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].

2. Implementation

(1) Binary Search

分别找左边界和右边界

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1, -1};

        if (nums == null || nums.length == 0) {
            return res;
        }

        int start = 0, end = nums.length - 1, mid = 0;

        while (start + 1 < end) {
            mid = start + (end - start) / 2;

            if (nums[mid] < target) {
                start = mid + 1;
            }
            else {
                end = mid;
            }
        }

        if (nums[start] == target) {
            res[0] = start;
        }
        else if (nums[end] == target) {
            res[0] = end;
        }
        else {
            return res;
        }

        end = nums.length - 1;

        while (start + 1 < end) {
            mid = start + (end - start) / 2;

            if (nums[mid] > target) {
                end = mid - 1;
            }
            else {
                start = mid;
            }
        }

        if (nums[end] == target) {
            res[1] = end;
        }
        else if (nums[start] == target) {
            res[1] = start;
        }
        else {
            return res;
        }
        return res;
    }
}

3. Time & Space Complexity

Binary Search: 时间复杂度O(logn), 空间复杂度O(1)