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## [0732. 我的日程安排表 III](https://leetcode.cn/problems/my-calendar-iii/) | ||
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- 标签:设计、线段树、有序集合 | ||
- 难度:困难 | ||
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## 题目大意 | ||
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**要求**:实现一个 `MyCalendarThree` 类来存放你的日程安排,你可以一直添加新的日程安排。 | ||
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日程可以用一对整数 $start$ 和 $end$ 表示,这里的时间是半开区间,即 $[start, end)$,实数 $x$ 的范围为 $start \le x < end$。 | ||
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`MyCalendarThree` 类: | ||
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- `MyCalendarThree()` 初始化对象。 | ||
- `int book(int start, int end)` 返回一个整数 `k`,表示日历中存在的 `k` 次预订的最大值。 | ||
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**说明**: | ||
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- `k` 次预定:当 `k` 个日程安排有一些时间上的交叉时(例如 `k` 个日程安排都在同一时间内),就会产生 `k` 次预订。 | ||
- $0 \le start < end \le 10^9$ | ||
- 每个测试用例,调用 `book` 函数最多不超过 `400` 次。 | ||
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**示例**: | ||
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```Python | ||
输入 | ||
["MyCalendarThree", "book", "book", "book", "book", "book", "book"] | ||
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]] | ||
输出 | ||
[null, 1, 1, 2, 3, 3, 3] | ||
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解释 | ||
MyCalendarThree myCalendarThree = new MyCalendarThree(); | ||
myCalendarThree.book(10, 20); // 返回 1 ,第一个日程安排可以预订并且不存在相交,所以最大 k 次预订是 1 次预订。 | ||
myCalendarThree.book(50, 60); // 返回 1 ,第二个日程安排可以预订并且不存在相交,所以最大 k 次预订是 1 次预订。 | ||
myCalendarThree.book(10, 40); // 返回 2 ,第三个日程安排 [10, 40) 与第一个日程安排相交,所以最大 k 次预订是 2 次预订。 | ||
myCalendarThree.book(5, 15); // 返回 3 ,剩下的日程安排的最大 k 次预订是 3 次预订。 | ||
myCalendarThree.book(5, 10); // 返回 3 | ||
myCalendarThree.book(25, 55); // 返回 3 | ||
``` | ||
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## 解题思路 | ||
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### 思路 1:线段树 | ||
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这道题可以使用线段树来做。 | ||
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因为区间的范围是 $[0, 10^9]$,普通数组构成的线段树不满足要求。需要用到动态开点线段树。 | ||
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- 构建一棵线段树。每个线段树的节点类存储当前区间中保存的日程区间个数。 | ||
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- 在 `book` 方法中,在线段树中更新 `[start, end - 1]` 的交叉日程区间个数,即令其区间值整体加 `1`。 | ||
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- 然后从线段树中查询区间 $[0, 10^9]$ 上保存的交叉日程区间个数,并返回。 | ||
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### 思路 1:代码 | ||
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```Python | ||
# 线段树的节点类 | ||
class SegTreeNode: | ||
def __init__(self, left=-1, right=-1, val=0, lazy_tag=None, leftNode=None, rightNode=None): | ||
self.left = left # 区间左边界 | ||
self.right = right # 区间右边界 | ||
self.mid = left + (right - left) // 2 | ||
self.leftNode = leftNode # 区间左节点 | ||
self.rightNode = rightNode # 区间右节点 | ||
self.val = val # 节点值(区间值) | ||
self.lazy_tag = lazy_tag # 区间问题的延迟更新标记 | ||
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# 线段树类 | ||
class SegmentTree: | ||
# 初始化线段树接口 | ||
def __init__(self, function): | ||
self.tree = SegTreeNode(0, int(1e9)) | ||
self.function = function # function 是一个函数,左右区间的聚合方法 | ||
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# 单点更新,将 nums[i] 更改为 val | ||
def update_point(self, i, val): | ||
self.__update_point(i, val, self.tree) | ||
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# 区间更新,将区间为 [q_left, q_right] 上的元素值修改为 val | ||
def update_interval(self, q_left, q_right, val): | ||
self.__update_interval(q_left, q_right, val, self.tree) | ||
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# 区间查询,查询区间为 [q_left, q_right] 的区间值 | ||
def query_interval(self, q_left, q_right): | ||
return self.__query_interval(q_left, q_right, self.tree) | ||
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# 获取 nums 数组接口:返回 nums 数组 | ||
def get_nums(self, length): | ||
nums = [0 for _ in range(length)] | ||
for i in range(length): | ||
nums[i] = self.query_interval(i, i) | ||
return nums | ||
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# 以下为内部实现方法 | ||
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# 单点更新,将 nums[i] 更改为 val。node 节点的区间为 [node.left, node.right] | ||
def __update_point(self, i, val, node): | ||
if node.left == node.right: | ||
node.val = val # 叶子节点,节点值修改为 val | ||
return | ||
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if i <= node.mid: # 在左子树中更新节点值 | ||
self.__update_point(i, val, node.leftNode) | ||
else: # 在右子树中更新节点值 | ||
self.__update_point(i, val, node.rightNode) | ||
self.__pushup(node) # 向上更新节点的区间值 | ||
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# 区间更新 | ||
def __update_interval(self, q_left, q_right, val, node): | ||
if node.left >= q_left and node.right <= q_right: # 节点所在区间被 [q_left, q_right] 所覆盖 | ||
if node.lazy_tag is not None: | ||
node.lazy_tag += val # 将当前节点的延迟标记增加 val | ||
else: | ||
node.lazy_tag = val # 将当前节点的延迟标记增加 val | ||
node.val += val # 当前节点所在区间增加 val | ||
return | ||
if node.right < q_left or node.left > q_right: # 节点所在区间与 [q_left, q_right] 无关 | ||
return 0 | ||
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self.__pushdown(node) # 向下更新节点所在区间的左右子节点的值和懒惰标记 | ||
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if q_left <= node.mid: # 在左子树中更新区间值 | ||
self.__update_interval(q_left, q_right, val, node.leftNode) | ||
if q_right > node.mid: # 在右子树中更新区间值 | ||
self.__update_interval(q_left, q_right, val, node.rightNode) | ||
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self.__pushup(node) | ||
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# 区间查询,在线段树的 [left, right] 区间范围中搜索区间为 [q_left, q_right] 的区间值 | ||
def __query_interval(self, q_left, q_right, node): | ||
if node.left >= q_left and node.right <= q_right: # 节点所在区间被 [q_left, q_right] 所覆盖 | ||
return node.val # 直接返回节点值 | ||
if node.right < q_left or node.left > q_right: # 节点所在区间与 [q_left, q_right] 无关 | ||
return 0 | ||
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self.__pushdown(node) # 向下更新节点所在区间的左右子节点的值和懒惰标记 | ||
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res_left = 0 # 左子树查询结果 | ||
res_right = 0 # 右子树查询结果 | ||
if q_left <= node.mid: # 在左子树中查询 | ||
res_left = self.__query_interval(q_left, q_right, node.leftNode) | ||
if q_right > node.mid: # 在右子树中查询 | ||
res_right = self.__query_interval(q_left, q_right, node.rightNode) | ||
return self.function(res_left, res_right) # 返回左右子树元素值的聚合计算结果 | ||
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# 向上更新 node 节点区间值,节点的区间值等于该节点左右子节点元素值的聚合计算结果 | ||
def __pushup(self, node): | ||
if node.leftNode and node.rightNode: | ||
node.val = self.function(node.leftNode.val, node.rightNode.val) | ||
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# 向下更新 node 节点所在区间的左右子节点的值和懒惰标记 | ||
def __pushdown(self, node): | ||
if node.leftNode is None: | ||
node.leftNode = SegTreeNode(node.left, node.mid) | ||
if node.rightNode is None: | ||
node.rightNode = SegTreeNode(node.mid + 1, node.right) | ||
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lazy_tag = node.lazy_tag | ||
if node.lazy_tag is None: | ||
return | ||
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if node.leftNode.lazy_tag is not None: | ||
node.leftNode.lazy_tag += lazy_tag # 更新左子节点懒惰标记 | ||
else: | ||
node.leftNode.lazy_tag = lazy_tag # 更新左子节点懒惰标记 | ||
node.leftNode.val += lazy_tag # 左子节点区间增加 lazy_tag | ||
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if node.rightNode.lazy_tag is not None: | ||
node.rightNode.lazy_tag += lazy_tag # 更新右子节点懒惰标记 | ||
else: | ||
node.rightNode.lazy_tag = lazy_tag # 更新右子节点懒惰标记 | ||
node.rightNode.val += lazy_tag # 右子节点区间增加 lazy_tag | ||
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node.lazy_tag = None # 更新当前节点的懒惰标记 | ||
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class MyCalendarThree: | ||
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def __init__(self): | ||
self.STree = SegmentTree(lambda x, y: max(x, y)) | ||
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def book(self, start: int, end: int) -> int: | ||
self.STree.update_interval(start, end - 1, 1) | ||
return self.STree.query_interval(0, int(1e9)) | ||
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# Your MyCalendarThree object will be instantiated and called as such: | ||
# obj = MyCalendarThree() | ||
# param_1 = obj.book(start,end) | ||
``` |