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最大栈

作者:互联网

题目:

思路:

​ 使用一个主栈(stack)+辅助栈(assistStack),assistStack 的栈顶永远是最大值。

1、push:

1.1 对于 stack 来说,直接push即可:stack.push(x);

1.2 对于 assistStack 来说,要进行判断,它 push 待插入的 x 和它栈顶两者最大的那个;

2、pop:两个栈都 pop,返回 stak 的 pop 值;

3、top:直接返回 stack 的 top 即可;

4、peekMax:assistStack 的栈顶保存的就是 stack 的最大值,于是:return assistStack.peek();

5、popMax:定义一个临时栈 tmpStack,判断 stack 的栈顶此时是不是最大值:

如果 stack.peak() != assistStack.peak(),说明不是最大值,把 stack 栈顶元素弹出并装入 tmpStack 中;

5.2 如果 stack.peak() == assistStack.peak(),说明是最大值,返回,并把 tmpStack 元素装回 stack 中。

代码:

class MaxStack {

    /** initialize your data structure here. */
    private final Stack<Integer> stack;
    private final Stack<Integer> assistStack;

    public MaxStack() {
        stack = new Stack<>();
        assistStack = new Stack<>();
    }

    public void push(int x) {
        stack.push(x);
        if (assistStack.isEmpty()) {
            assistStack.push(x);
        } else {
            if (assistStack.peek() < x) {
                assistStack.push(x);
            } else {
                assistStack.push(assistStack.peek());
            }
        }
    }

    public int pop() {
        assistStack.pop();
        return stack.pop();
    }

    public int top() {
        return stack.peek();
    }

    public int peekMax() {
        return assistStack.peek();
    }

    public int popMax() {
        Stack<Integer> tmpStack = new Stack<>();
        while (!stack.peek().equals(assistStack.peek())) {
            tmpStack.push(stack.pop());
        }
        int max = stack.pop();
        while (!tmpStack.isEmpty()) {
            stack.push(tmpStack.pop());
        }
        return max;
    }
}

/**
 * Your MaxStack object will be instantiated and called as such:
 * MaxStack obj = new MaxStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.peekMax();
 * int param_5 = obj.popMax();
 */

标签:peek,最大,int,pop,push,assistStack,stack
来源: https://www.cnblogs.com/ArtiaDeng-blog/p/16244433.html