Search for a Range

Search for a Range

 

Given a sorted array of integers, 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]

.

For example,

Given 

[5, 7, 7, 8, 8, 10]

 and target value 8,

return 

[3, 4]

.

use binary search to search for the ranges. 

public class Solution {
public int[] searchRange(int[] nums, int target) {
if (nums == null || nums.length == 0)
return new int[]{-1,-1};

int[] bound = new int[2];

int start = 0;
int end = nums.length - 1;
int mid = 0;
while (start < end - 1) {
mid = start + (end - start) / 2;
if (nums[mid] == target)
end = mid;

else if (nums[mid] > target)
end = mid;

else
start = mid;

}

if (nums[start] == target)
bound[0] = start;

else if (nums[end] == target)
bound[0] = end;

else {
return new int[]{-1,-1};
}

start = 0;
end = nums.length - 1;

while (start < end - 1) {
mid = start + (end - start) / 2;
if (nums[mid] == target)
start = mid;

else if (nums[mid] > target)
end = mid;

else
start = mid;

}

if (nums[end] == target)
bound[1] = end;

else if (nums[start] == target)
bound[1] = start;

else {
return new int[]{-1,-1};
}

return bound;

}
}