LeetCode 155. Min Stack 返回栈中最小值

Question:

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:

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.

原题链接：https://leetcode.com/problems/min-stack/description/

思路：

利用两个栈，一个存放数据data，一个存放当前栈的最小值min。每次数据栈data压入新数据x，都判断新数据x是不是当前最小，是的话，将x放入最小值栈min,不是的话，将min的顶部值再压入一次。每次data数据出栈，min也要出栈一个值，即最小值栈与数据栈保持同步更新。时间复杂度为O(1)，用空间换时间。

class MinStack {

    /** initialize your data structure here. */
    public MinStack() {
        
    }
    private Stack  data = new Stack  ();
    private Stack  min = new Stack  ();
    
    public void push(int x) {
        if (data.empty())
        {  
            data.push(x);
            min.push(x);
         }
        else
        {
            if(x < min.peek())
            {
                data.push(x);
                min.push(x);
            }
            else
            {
                data.push(x);
                min.push(min.peek());
            }
        }
    }
    
    public void pop() {
        data.pop();
        min.pop();
    }
    
    public int top() {
        return data.peek();
    }
    
    public int getMin() {
        return min.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();
 */

博主原创，未经博主允许，不得转载，谢谢~如果有错误请提醒我^-^