LeetCode 34 - 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]

.

My Code

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int size = nums.size();
int left_idx = 0;
int right_idx = size - 1;
int mid_idx;
vector<int> range(2, -1);

// Find start
while (left_idx <= right_idx)
{
mid_idx = (left_idx + right_idx) / 2;
if (nums[mid_idx] == target)
{
if (mid_idx == 0 || nums[mid_idx-1] < target)
{
range[0] = mid_idx;
break;
}
else
right_idx = mid_idx - 1;
}
else if (nums[mid_idx] < target)
left_idx = mid_idx + 1;
else
right_idx = mid_idx - 1;
}

// Find end
if (range[0] != -1)
{
left_idx = range[0] + 1;
right_idx = size - 1;

while (left_idx <= right_idx)
{
mid_idx = (left_idx + right_idx) / 2;
if (nums[mid_idx] == target)
{
if (mid_idx == size - 1 || nums[mid_idx+1] > target)
{
range[1] = mid_idx;
break;
}
else
left_idx = mid_idx + 1;
}
else if (nums[mid_idx] < target)
left_idx = mid_idx + 1;
else
right_idx = mid_idx - 1;
}
}

if (range[1] == -1)
range[1] = range[0];

return range;
}
};

Runtime: 12 ms