【Leetcode-Easy-155】 Min Stack

题目

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.

思路

双栈

程序

class MinStack {    private LinkedList<Integer> dataStack = null;    private LinkedList<Integer> minStack = null;     /** initialize your data structure here. */    public MinStack() {        dataStack = new LinkedList<>();        minStack = new LinkedList<>();    }    public void push(int x) {        dataStack.push(x);        if (minStack.isEmpty()) minStack.push(x);        else                    minStack.push(Math.min(minStack.peek(), x));    }    public void pop() {        if(!minStack.isEmpty()) minStack.pop();        if(!dataStack.isEmpty()) dataStack.pop();    }    public int top() {        return dataStack.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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声明：题目来自Leetcode.