Contains Duplicate III

Given an array of integers, find out whether there are two distinct indices i and j in
the array such that the difference between nums[i] and nums[j] is
at most t and
the difference between i and j is
at most k.

class Solution
{
public:
bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t)
{
if (nums.size() < 2 || k == 0)
return false;
deque<int> windows_deq;
multiset<long> windows;
//维护一个长度小于等于k的滑动窗口
for (int i = 0; i < nums.size(); i++)
{
if (windows.size() > k)
{
int num = windows_deq.front();
windows_deq.pop_front();
windows.erase(windows.find(num));
}

//这句很关键，num[i]与num[j]的差值最大为t，也就是说num[j]的最小值为num[i]-t
auto it = windows.lower_bound((long)nums[i] - (long)t);

if (it == windows.end() || *it > (long)nums[i] + (long)t) {
// not found
windows_deq.push_back(nums[i]);
windows.insert(nums[i]);
}
else
return true;
}
return false;
}
};