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FirstAndLastPosition.java
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/*34. Find First and Last Position of Element in Sorted Array --> https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/description/ (Medium)
* Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.
If target is not found in the array, return [-1, -1].
You must write an algorithm with O(log n) runtime complexity.
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
*
*
*
*
*/
public class FirstAndLastPosition
{
public static int[] searchRange(int[] nums, int target)
{
int[] result = {-1, -1};
// Search for the leftmost occurrence
result[0] = binarySearch(nums, target, true);
// If the leftmost occurrence is not found, the rightmost won't be either
if (result[0] == -1)
{
return result;
}
// Search for the rightmost occurrence
result[1] = binarySearch(nums, target, false);
return result;
}
private static int binarySearch(int[] nums, int target, boolean findLeftmost)
{
int start = 0;
int end = nums.length - 1;
int result = -1;
while (start <= end)
{
int mid = start + (end - start) / 2;
if (nums[mid] == target)
{
result = mid;
// Adjust the search space based on whether we want to find the leftmost or rightmost occurrence
if (findLeftmost)
{
end = mid - 1;
}
else
{
start = mid + 1;
}
}
else if (nums[mid] < target)
{
start = mid + 1;
}
else
{
end = mid - 1;
}
}
return result;
}
public static void main(String[] args)
{
int[] nums = {5, 7, 7, 8, 8, 10};
int target = 8;
int[] result = searchRange(nums, target);
System.out.println("Starting Position: " + result[0]);
System.out.println("Ending Position: " + result[1]);
}
}