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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/dui-lie-de-zui-da-zhi-lcof
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
代码:
class MaxQueue { public: queue<int> q; deque<int> p; MaxQueue() { } int max_value() { if(p.empty()) return -1; return p.front(); } void push_back(int value) { q.push(value); while(!p.empty() && p.back() < value) p.pop_back(); p.push_back(value); } int pop_front() { if(q.empty()) return -1; if(p.front() == q.front()) p.pop_front(); int d = q.front(); q.pop(); return d; } }; /** * 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(); */