力扣 LeetCode - 剑指 Offer 59 - II. 队列的最大值

题目：

请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都是O(1)。 若队列为空，pop_front 和 max_value 需要返回 -1 示例 1： 输入: ["MaxQueue","push_back","push_back","max_value","pop_front","max_value"] [[],[1],[2],[],[],[]] 输出: [null,null,null,2,1,2] 示例 2： 输入: ["MaxQueue","pop_front","max_value"] [[],[],[]] 输出: [null,-1,-1] 限制： 1 <= push_back,pop_front,max_value的总操作数 <= 10000 1 <= value <= 10^5

题目解析：

JAVA：

class MaxQueue {
	List<Integer> list;
	Integer maxIndex = 0;

	public MaxQueue() {
		list = new ArrayList<>();
	}

	public int max_value() {
		if (list.size() == 0) {
			return -1;
		}
		if (maxIndex == 0 && list.size() > 0) {
			Integer maxIndexTemp = -1;
			for (int i = 0; i < list.size(); i++) {
				if (maxIndexTemp == -1 || list.get(i) > list.get(maxIndexTemp)) {
					maxIndexTemp = i;
				}
			}
			if (maxIndexTemp == -1) {
				return -1;
			}
			maxIndex = maxIndexTemp;
			return list.get(maxIndex);
		}
		return list.get(maxIndex);
	}

	public void push_back(int value) {
		list.add(value);
		if (maxIndex > 0 && value > list.get(maxIndex)) {
			maxIndex = list.size() - 1;
		}
	}

	public int pop_front() {
		if (list.size() == 0) {
			return -1;
		} else {
			if (maxIndex > 0) {
				maxIndex = maxIndex - 1;
			}
			return list.remove(0);
		}
	}
}

/**
 * Your MaxQueue object will be instantiated and called as such:
 * MaxQueue obj = new MaxQueue();
 * int param_1 = obj.max_value();
 * obj.push_back(value);
 * int param_3 = obj.pop_front();
 */