大年夜功课请求写个中缀表达式转后缀表达式，结不雅发明视频琅绫擎给的解法是个错的，网上找了一大年夜堆，各类说法都有，公说公有理婆说婆有理，花里胡哨的，看着也复杂的一匹，说的话旁敲侧击的，看几下就开端看不懂了，研究到凌晨终于画了一张：秒杀大年夜图，图是基于调剂场算法（Shunting-Yard algorithm）的流程图，可以说是清楚清楚明了，就照着图无脑撸代码就完事了

代码如下（仅参考，建议照样看图本身敲的舒畅）：

	public Queue infixToPostfix(Queue infix) {
		Stack stack = new Stack();
		Queue postfix = new Queue();
		Boolean backtoRoot = false;
		while (!infix.isEmpty()){
			backtoRoot = false;

			Data data = infix.poll();

			if (data.type == Type.OPERATOR || data.type == Type.PAREN){
				if ("+-*/".contains(data.getOperator() )){
					while (!stack.isEmpty()) { //Stack is not empty
						Token top = stack.peek();
						if (("+-(".contains(top.getOperator()) & "/*".contains(data.getOperator())) | ((top.getOperator().equals("(")) & "-+".contains(data.getOperator() ))){ //栈顶元素优先级低于本身
							stack.push(data);
							backtoRoot = true;
							break;
						}else {
							postfix.offer(stack.pop());
							continue;
						}

					}
					if (backtoRoot){
						continue;
					}
					// Stack is empty
					if (stack.isEmpty()){
						stack.push(data);
						continue;
					}
					continue;

				}
				if ((data.getOperator().equals("(") )){
					stack.push(data);
					continue;
				}
				if (")".contains(data.getOperator() )){
					while (!stack.isEmpty() ){

						if (stack.peek().getOperator().equals("(")) {
							stack.pop();
							break;
						}
						postfix.offer(stack.pop());

					}
					continue;
				}
			}

			if (data.type == Type.OPERAND){
				postfix.offer(data);
				continue;
			}

		}


		while (!stack.isEmpty()){
			postfix.offer(stack.pop());
		}
		return postfix;
	}