英文:
Why can't I remove an element gotten by peek() from the PriorityQueue?
问题
以下是翻译好的代码部分：
class MinStack {
public Deque<Integer> deque = new LinkedList<Integer>();
public PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
public MinStack() {
Deque<Integer> deque = new LinkedList<Integer>();
PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
}
public void push(int x) {
deque.offer(x);
pq.offer(x);
}
public void pop() {
pq.remove(deque.peek());
deque.pollLast();
}
public int top() {
return deque.peekLast();
}
public int getMin() {
return pq.peek();
}
}
在函数pop()中，优先队列PriorityQueue不会删除我从deque.peek()获取的顶部值。当我将其更改为
pq.remove(deque.pollLast());
它就正常工作了。为什么会这样呢？
英文:
Here is my code.
class MinStack {
public Deque<Integer> deque = new LinkedList<Integer>();
public PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
public MinStack() {
Deque<Integer> deque = new LinkedList<Integer>();
PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
}
public void push(int x) {
deque.offer(x);
pq.offer(x);
}
public void pop() {
pq.remove(deque.peek());
deque.pollLast();
}
public int top() {
return deque.peekLast();
}
public int getMin() {
return pq.peek();
}
}
In the function pop(), the PriorityQueue doesn't delete the top value I get from the deque.peek().
When I changed it to
pq.remove(deque.pollLast());
It worked. Why is that?
答案1
得分: 2
Deque.peek()返回双端队列的第一个元素，与peekFirst()相同。在top()中所做的那样，改用peekLast()。
英文:
Deque.peek() returns the first element of the deque, same as peekFirst(). Use peekLast() instead like you did in top().
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论