剑指 Offer 59 - II. 队列的最大值 做题小结
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
题解:
代码:
//方法一:
class MaxQueue {
int[] q = new int[20000];
int begin = 0, end = 0;
public MaxQueue() {
}
public int max_value() {
int ans = -1;
for (int i = begin; i != end; ++i) {
ans = Math.max(ans, q[i]);
}
return ans;
}
public void push_back(int value) {
q[end++] = value;
}
public int pop_front() {
if (begin == end) {
return -1;
}
return q[begin++];
}
}
//方法二
class MaxQueue {
Queue<Integer> q;
Deque<Integer> d;
public MaxQueue() {
q = new LinkedList<Integer>();
d = new LinkedList<Integer>();
}
public int max_value() {
if (d.isEmpty()) {
return -1;
}
return d.peekFirst();
}
public void push_back(int value) {
while (!d.isEmpty() && d.peekLast() < value) {
d.pollLast();
}
d.offerLast(value);
q.offer(value);
}
public int pop_front() {
if (q.isEmpty()) {
return -1;
}
int ans = q.poll();
if (ans == d.peekFirst()) {
d.pollFirst();
}
return ans;
}
}