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很簡(jiǎn)單只要記住add操作不需要倒棧 poll 和peek的時(shí)候判斷性的倒棧 肯定能設(shè)計(jì)出來
package com.offer.chapter1;import java.util.Stack;/** * CopyRright (c)2014-2016 Haerbin Hearglobal Co.,Ltd * PRoject: demo * Comments: * Author:cbam * Create Date:2017/2/4 * Modified By: * Modified Date: * Modified Reason: */public class Problem_02_TwoStacksQueue { private Stack<Integer> stackPush; private Stack<Integer> stackPop; public Problem_02_TwoStacksQueue() { stackPush = new Stack<>(); stackPop = new Stack<>(); } public void add(int pushInt) { stackPush.push(pushInt); } public int poll() { if(stackPop.empty() && stackPush.empty()) { throw new RuntimeException("Queue is empty"); } else if(stackPop.empty()) { while(!stackPush.empty()) { stackPop.push(stackPush.pop()); } } return stackPop.pop(); } public int peek() { if(stackPop.empty() && stackPush.empty()) { throw new RuntimeException("Queue is empty"); } else if(stackPop.empty()) { while(!stackPush.empty()) { stackPop.push(stackPush.pop()); } } return this.stackPop.peek(); }}新聞熱點(diǎn)
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