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FirstAndLastPositionOfAnElement.java
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FirstAndLastPositionOfAnElement.java
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package Leetcode;
/**
* @author kalpak
*
* Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.
*
* If target is not found in the array, return [-1, -1].
*
* Follow up: Could you write an algorithm with O(log n) runtime complexity?
*
*
*
* Example 1:
*
* Input: nums = [5,7,7,8,8,10], target = 8
* Output: [3,4]
* Example 2:
*
* Input: nums = [5,7,7,8,8,10], target = 6
* Output: [-1,-1]
* Example 3:
*
* Input: nums = [], target = 0
* Output: [-1,-1]
*
*
* Constraints:
*
* 0 <= nums.length <= 10^5
* -10^9 <= nums[i] <= 10^9
* nums is a non-decreasing array.
* -10^9 <= target <= 10^9
*/
public class FirstAndLastPositionOfAnElement {
public static int[] searchRange(int[] nums, int target) {
if(nums == null || nums.length == 0)
return new int[]{-1, -1};
int[] result = new int[2];
result[0] = -1;
result[1] = -1;
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left)/2;
if (nums[mid] == target) {
result[0] = mid;
right = mid - 1;
} else if (nums[mid] < target)
left = mid + 1;
else
right = mid - 1;
}
left = 0;
right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left)/2;
if (nums[mid] == target) {
result[1] = mid;
left = mid + 1;
} else if (nums[mid] < target)
left = mid + 1;
else
right = mid - 1;
}
return result;
}
public static void main(String[] args) {
int[] arr = new int[]{5,7,7,8,8,10};
int[] result = searchRange(arr, 8);
System.out.println("["+result[0]+", "+result[1]+"]" );
}
}