【LeetCode-Java】155. Min Stack

1.原题

链接：https://leetcode.com/problems/min-stack/

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.

2.题目大意

定义栈的数据结构，实现一个能够得到栈的最小元素的min函数。在该栈中，调用min、push及pop的时间复杂度都是O(1)

3.解题思路

为了能在O(1)时间内得到最小元素，需要有辅助的栈，保存当前的最小值，记为最小栈。这个辅助的栈，需在push(),pop()操作时更新，push时若当前最小栈为空或入栈的值<=最小栈栈顶的值，则将当前入栈的值压入最小栈;pop时，若出栈的值和最小栈栈顶的值相同，则最小栈也pop。

4.代码实现

public class MinStack {
ArrayList<Integer> stack=new ArrayList<>();
ArrayList<Integer> minStack=new ArrayList<>();  
    public void push(int x) {
    stack.add(x);
    if(minStack.isEmpty() || minStack.get(minStack.size()-1)>=x)
    minStack.add(x);  
    }   
    public void pop() {
    if(stack.isEmpty())
    return;
    int last=stack.remove(stack.size()-1);
    if(!minStack.isEmpty() && minStack.get(minStack.size()-1)==last)
    minStack.remove(minStack.size()-1);     
    } 
    public int top() {
    if(stack.isEmpty())
    return 0;
    return stack.get(stack.size()-1);   
    }
    public int getMin() {
    if(minStack.isEmpty())
    return 0;
    return minStack.get(minStack.size()-1);
    }
}