练习2
实现一个特殊的栈,在实现栈的基本功能的基础上,再实现返 回栈中最小元素的操作。
【要求】
1.pop、push、getMin操作的时间复杂度都是O(1)。
2.设计的栈类型可以使用现成的栈结构。
思路:
同时维护俩个栈
1)如果新增的数比min栈顶小,将该数压入min栈和data栈;如果新增的数比min栈顶大,将该数压入data栈,同时将重新压入min栈顶上的数
练习3
如何仅用队列结构实现栈结构? 如何仅用栈结构实现队列结构?
public static class MyStack1 { private Stack<Integer> stackData; private Stack<Integer> stackMin; public MyStack1() { this.stackData = new Stack<Integer>(); this.stackMin = new Stack<Integer>(); } public void push(int newNum) { if (this.stackMin.isEmpty()) { this.stackMin.push(newNum); } else if (newNum <= this.getmin()) { this.stackMin.push(newNum); } this.stackData.push(newNum); } public int pop() { if (this.stackData.isEmpty()) { throw new RuntimeException("Your stack is empty."); } int value = this.stackData.pop(); if (value == this.getmin()) { this.stackMin.pop(); } return value; } public int getmin() { if (this.stackMin.isEmpty()) { throw new RuntimeException("Your stack is empty."); } return this.stackMin.peek(); } } public static class MyStack2 { private Stack<Integer> stackData; private Stack<Integer> stackMin; public MyStack2() { this.stackData = new Stack<Integer>(); this.stackMin = new Stack<Integer>(); } public void push(int newNum) { if (this.stackMin.isEmpty()) { this.stackMin.push(newNum); } else if (newNum < this.getmin()) { this.stackMin.push(newNum); } else { int newMin = this.stackMin.peek(); this.stackMin.push(newMin); } this.stackData.push(newNum); } public int pop() { if (this.stackData.isEmpty()) { throw new RuntimeException("Your stack is empty."); } this.stackMin.pop(); return this.stackData.pop(); } public int getmin() { if (this.stackMin.isEmpty()) { throw new RuntimeException("Your stack is empty."); } return this.stackMin.peek(); } }