LeetCode_155.最小栈

设计一个支持 pushpoptop 操作,并能在常数时间内检索到最小元素的栈。

  • push(x) —— 将元素 x 推入栈中。
  • pop() —— 删除栈顶的元素。
  • top() —— 获取栈顶元素。
  • getMin() —— 检索栈中的最小元素。

 

示例:

输入:
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

输出:
[null,null,null,null,-3,null,0,-2]

解释:
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.

 

提示:

  • poptopgetMin 操作总是在 非空栈 上调用。

C#代码

public class MinStack {

    private Stack<Tuple<int, int>> stack;
    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack<Tuple<int, int>>();
    }
    
    public void Push(int x) {
        var min=x;
        if(stack.Any()){
            var top = stack.Peek();
            min = Math.Min(top.Item2,min);
        }
        stack.Push(Tuple.Create(x,min));
    }
    
    public void Pop() {
        stack.Pop();
    }
    
    public int Top() {
        var top = stack.Peek();
        return top.Item1;
    }
    
    public int GetMin() {
        var top = stack.Peek();
        return top.Item2;
    }
}

/**
 * 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();
 */
posted @ 2021-01-06 18:34  付旭洋  阅读(61)  评论(0编辑  收藏  举报