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# Implement Queue by Two Stacks ### Source - lintcode: [(40) Implement Queue by Two Stacks](http://www.lintcode.com/en/problem/implement-queue-by-two-stacks/) ~~~ As the title described, you should only use two stacks to implement a queue's actions. The queue should support push(element), pop() and top() where pop is pop the first(a.k.a front) element in the queue. Both pop and top methods should return the value of first element. Example For push(1), pop(), push(2), push(3), top(), pop(), you should return 1, 2 and 2 Challenge implement it by two stacks, do not use any other data structure and push, pop and top should be O(1) by AVERAGE. ~~~ ### 題解 兩個(gè)棧模擬隊(duì)列，棧是 LIFO, 隊(duì)列是 FIFO, 故用兩個(gè)棧模擬隊(duì)列時(shí)可結(jié)合棧1和棧2, LIFO + LIFO ==> FIFO, 即先將一個(gè)棧元素全部 push 到另一個(gè)棧，效果即等價(jià)于 Queue. ### Java ~~~ public class Solution { private Stack<Integer> stack1; private Stack<Integer> stack2; public Solution() { // source stack stack1 = new Stack<Integer>(); // target stack stack2 = new Stack<Integer>(); } public void push(int element) { stack1.push(element); } public int pop() { if (stack2.empty()) { stack1ToStack2(stack1, stack2); } return stack2.pop(); } public int top() { if (stack2.empty()) { stack1ToStack2(stack1, stack2); } return stack2.peek(); } private void stack1ToStack2(Stack<Integer> stack1, Stack<Integer> stack2) { while (!stack1.empty()) { stack2.push(stack1.pop()); } } } ~~~ ### 源碼分析 將棧1作為原始棧，將棧1元素壓入棧2是公共方法，故寫成一個(gè)私有方法。 ### 復(fù)雜度分析 視連續(xù) push 的元素而定，時(shí)間復(fù)雜度近似為 O(1)O(1)O(1). ### Reference - [Implement Queue by Two Stacks 參考程序 Java/C++/Python](http://www.jiuzhang.com/solutions/implement-queue-by-two-stacks/)