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]

.

使用二分查找先查找左边界，然后查找右边界

注意：题目假设target value一定可以找到，所以，不需要判断标记是否出界

vector<int> searchRange(vector<int>& nums, int target) {
int beg = 0, end = nums.size() - 1, mid = 0;
vector<int> ret = {-1, -1};
// 查找左边界
while (beg <= end)
{
mid = beg + ((end - beg) >> 1);
if (target <= nums[mid])
end = mid - 1;
else
beg = mid + 1;
}

if (nums[beg] == target)
ret[0] = beg;
else
return ret;
// 查找右边界
end = nums.size() - 1; // 从左边界开始查找
while (beg <= end)
{
mid = beg + ((end - beg) >> 1);
if (target < nums[mid])
end = mid - 1;
else
beg = mid + 1;
}
ret[1] = end;
return ret;
}