Simple lower-bound & upper-bound solution in Java, cost 0ms

• The solution uses two auxiliary functions, lower_bound and upper_bound, which find smallest & largest indexes of the number @target in the array @nums respectively.

public class Solution {
private int lower_bound(int[] nums, int target) {
int low = 0, high = nums.length - 1;
while(low < high) {
int mid = low + (high - low) / 2;
if(nums[mid] < target)
low = mid + 1;
else
high = mid;
}
return nums[low] == target? low : -1;
}
private int upper_bound(int[] nums, int target) {
int low = 0, high = nums.length - 1;
while(low < high) {
int mid = low + (high+1 - low) / 2;
if(nums[mid] > target)
high = mid - 1;
else
low = mid;
}
return nums[low] == target? low : -1;
}
public int[] searchRange(int[] nums, int target) {
int[] result = new int[2];
result[0] = lower_bound(nums, target);
result[1] = upper_bound(nums, target);
return result;
}

}

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.