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栈的最小值

article 2024/9/30 21:09:28

请设计一个栈，除了常规栈支持的pop与push函数以外，还支持min函数，该函数返回栈元素中的最小值。执行push、pop和min操作的时间复杂度必须为O(1)。

示例：

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

class MinStack {
    private Stack<Integer> stack;
    private Stack<Integer> minStack;
    
    public MinStack() {
        this.stack = new Stack<>();
        this.minStack = new Stack<>();
    }
    
    public void push(int x) {
        stack.push(x);
        if(minStack.isEmpty() || minStack.peek()>=x){//等于是为了适应后面的出栈
            minStack.push(x);
        }
    }
    
    public void pop() {
        int n = stack.pop();
        if(n == minStack.peek()){//适应这里
            minStack.pop();
        }
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int getMin() {
        return minStack.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();
 */

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