Leetcode(155) Min Stack

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Description

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

  • push(x) – Push element x onto stack.
  • pop() – Removes the element on top of the stack.
  • top() – Get the top element.
  • getMin() – Retrieve the minimum element in the stack.

Example:

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MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

解法

用两个栈来实现，其中一个栈正常地存数，另一个存对饮栈状态第一个栈中的最小值即可。

具体代码如下：

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class MinStack {
    
    Stack<Integer> s;
    Stack<Integer> helper;
    /** initialize your data structure here. */
    public MinStack() {
        this.s = new Stack<>();
        this.helper = new Stack<>();
    }
    
    public void push(int x) {
        s.push(x);
        if(this.helper.size() == 0){
            this.helper.push(x);
        }
        else{
            if(this.helper.peek() < x){
                this.helper.push(this.helper.peek());
            }
            else{
                this.helper.push(x);
            }
        }
    }
    
    public void pop() {
       this.s.pop();
        this.helper.pop();
    }
    
    public int top() {
        return this.s.peek();
    }
    
    public int getMin() {
        return this.helper.peek();
    }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.getMin();
 */