直观思路是在增量操作时进行模拟,每次进行增量操作时都遍历指定的 k 个元素,但这样时间复杂度是 $O(N)$。
时间优化思路是,既然我们只有在出栈时才考虑元素的值,何不将增量操作延后到出栈时,如此便需要将所有增量操作都先存起来。
class CustomStack {
public:
CustomStack(int maxSize): max_size_(maxSize) {
stack_.resize(maxSize);
add_.resize(maxSize);
}
void push(int x) {
if (top_ < max_size_ - 1) {
top_++;
stack_[top_] = x;
}
}
int pop() {
if (top_ == -1) return -1;
int res = stack_[top_] + add_[top_];
if (top_ > 0) add_[top_ - 1] += add_[top_];
add_[top_] = 0;
top_--;
return res;
}
void increment(int k, int val) {
int idx = min(k - 1, top_);
if (idx >= 0) add_[idx] += val;
}
private:
vector<int> stack_;
vector<int> add_;
int top_ = -1;
int max_size_;
};
/**
* Your CustomStack object will be instantiated and called as such:
* CustomStack* obj = new CustomStack(maxSize);
* obj->push(x);
* int param_2 = obj->pop();
* obj->increment(k,val);
*/