Find First and Last Position of Element in Sorted Array
Given an array of integers nums
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]
.
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]
public class Solution {
public int[] searchRange(final int[] A, int B) {
// Binary searching for 1st occourence
int left = binarySearch(A, B, false);
// If we cannot find any occourence
if (left == -1)
return new int[] { -1, -1 };
// Find the last occourence
return new int[] { left, binarySearch(A, B, true) };
}
private int binarySearch(final int[] A, int target, boolean searchLast) {
int low = 0, high = A.length - 1;
int result = -1;
while (low <= high) {
int mid = (low + high) / 2;
if (A[mid] == target) {
result = mid;
if (searchLast)
low = mid + 1;
else
high = mid - 1;
} else if (target > A[mid])
low = mid + 1;
else
high = mid - 1;
}
return result;
}
}
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