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[LeetCode] 716. Max Stack

2022-03-20 07:02:22 作者：互联网

Design a max stack data structure that supports the stack operations and supports finding the stack's maximum element.

Implement the MaxStack class:

MaxStack() Initializes the stack object.
void push(int x) Pushes element x onto the stack.
int pop() Removes the element on top of the stack and returns it.
int top() Gets the element on the top of the stack without removing it.
int peekMax() Retrieves the maximum element in the stack without removing it.
int popMax() Retrieves the maximum element in the stack and removes it. If there is more than one maximum element, only remove the top-most one.

Example 1:

Input
["MaxStack", "push", "push", "push", "top", "popMax", "top", "peekMax", "pop", "top"]
[[], [5], [1], [5], [], [], [], [], [], []]
Output
[null, null, null, null, 5, 5, 1, 5, 1, 5]

Explanation
MaxStack stk = new MaxStack();
stk.push(5);   // [5] the top of the stack and the maximum number is 5.
stk.push(1);   // [5, 1] the top of the stack is 1, but the maximum is 5.
stk.push(5);   // [5, 1, 5] the top of the stack is 5, which is also the maximum, because it is the top most one.
stk.top();     // return 5, [5, 1, 5] the stack did not change.
stk.popMax();  // return 5, [5, 1] the stack is changed now, and the top is different from the max.
stk.top();     // return 1, [5, 1] the stack did not change.
stk.peekMax(); // return 5, [5, 1] the stack did not change.
stk.pop();     // return 1, [5] the top of the stack and the max element is now 5.
stk.top();     // return 5, [5] the stack did not change.

Constraints:

-107 <= x <= 107
At most 104 calls will be made to push, pop, top, peekMax, and popMax.
There will be at least one element in the stack when pop, top, peekMax, or popMax is called.

Follow up: Could you come up with a solution that supports O(1) for each top call and O(logn) for each other call?

最大栈。

设计一个支持push，pop，top，peekMax和popMax操作的最大栈。

push(x) -- 将元素x添加到栈中。
pop() -- 删除栈中最顶端的元素并将其返回。
top() -- 返回栈中最顶端的元素。
peekMax() -- 返回栈中最大的元素。
popMax() -- 返回栈中最大的元素，并将其删除。如果有多于一个最大的元素，只删除最靠近顶端的一个元素。

-1e7 <= x <= 1e7
操作的个数不会超过10000.
当栈为空时，后面四个操作不会被调用。

这是一道设计题，类似155题，但是比155题要难得多。

题意不难理解，思路如下，这里我们需要两个 stack，一个普通的栈 stack 存放所有需要处理的元素，另一个栈 maxStack 只存放当前遍历到的最大的元素，所以在任一时间点，maxStack 的顶端永远都是当前你处理过的元素中最大的那一个。

push() - 因为我们需要满足 maxStack 的栈顶永远是当前遍历到的最大元素，所以我们在做 push 操作的时候，对于每个元素 x，我们需要判断 x 是否大于当前的最大值 curMax，如果是，则更新 curMax 且将 curMax 放到 maxStack 中。这里 push() 操作的复杂度是 O(1)

pop() - 对两个栈都仅做 pop() 操作，复杂度是 O(1)

top() - 查看 stack 的顶端元素，复杂度是 O(1)

peekMax() - 查看 maxStack 的顶端元素，复杂度是 O(1)

popMax() - popMax() 的定义是返回栈中最大的元素，并将其删除。如果有多于一个最大的元素，只删除最靠近顶端的一个元素。因为 maxStack 的顶端一定是栈中最大的元素 max，所以对于 maxStack 而言，我们做一下 pop() 操作即可。但是对于 stack 而言，这个最大的元素不一定在 stack 的顶端，所以这里我们还需要创建一个临时的栈 temp 来存放从 stack 中弹出的所有元素，直到遇到 max 为止。从 stack 和 maxStack 中弹出 max 之后，我们再将 temp 中的元素再 push 回 stack 中，这样我们就做到了仅从 stack 中去除栈中最大的元素 max而不影响到其他元素的效果。这个操作的复杂度是 O(n)

Java实现

 1 class MaxStack {
 2     Stack<Integer> stack;
 3     Stack<Integer> maxStack;
 4 
 5     public MaxStack() {
 6         stack = new Stack<>();
 7         maxStack = new Stack<>();
 8     }
 9     
10     public void push(int x) {
11         helper(x);
12     }
13     
14     public void helper(int x) {
15         int curMax = maxStack.isEmpty() ? Integer.MIN_VALUE : maxStack.peek();
16         if (x > curMax) {
17             curMax = x;
18         }
19         stack.push(x);
20         maxStack.push(curMax);
21     }
22 
23     public int pop() {
24         maxStack.pop();
25         return stack.pop();
26     }
27     
28     public int top() {
29         return stack.peek();
30     }
31     
32     public int peekMax() {
33         return maxStack.peek();
34     }
35     
36     public int popMax() {
37         int max = maxStack.peek();
38         Stack<Integer> temp = new Stack<>();
39         while (stack.peek() != max) {
40             temp.push(stack.pop());
41             maxStack.pop();
42         }
43         stack.pop();
44         maxStack.pop();
45         while (!temp.isEmpty()) {
46             int x = temp.pop();
47             helper(x);
48         }
49         return max;
50     }
51 }
52 
53 /**
54  * Your MaxStack object will be instantiated and called as such:
55  * MaxStack obj = new MaxStack();
56  * obj.push(x);
57  * int param_2 = obj.pop();
58  * int param_3 = obj.top();
59  * int param_4 = obj.peekMax();
60  * int param_5 = obj.popMax();
61  */