描述
Given an empty set of intervals, implement a data structure that can:
- Add an interval to the set of intervals.
- Count the number of integers that are present in at least one interval.
Implement the CountIntervals class:
- CountIntervals() Initializes the object with an empty set of intervals.
- void add(int left, int right) Adds the interval [left, right] to the set of intervals.
- int count() Returns the number of integers that are present in at least one interval.
Note that an interval [left, right] denotes all the integers x where left <= x <= right.
Example 1:
Input
["CountIntervals", "add", "add", "count", "add", "count"]
[[], [2, 3], [7, 10], [], [5, 8], []]
Output
[null, null, null, 6, null, 8]
Explanation
CountIntervals countIntervals = new CountIntervals(); // initialize the object with an empty set of intervals.
countIntervals.add(2, 3); // add [2, 3] to the set of intervals.
countIntervals.add(7, 10); // add [7, 10] to the set of intervals.
countIntervals.count(); // return 6
// the integers 2 and 3 are present in the interval [2, 3].
// the integers 7, 8, 9, and 10 are present in the interval [7, 10].
countIntervals.add(5, 8); // add [5, 8] to the set of intervals.
countIntervals.count(); // return 8
// the integers 2 and 3 are present in the interval [2, 3].
// the integers 5 and 6 are present in the interval [5, 8].
// the integers 7 and 8 are present in the intervals [5, 8] and [7, 10].
// the integers 9 and 10 are present in the interval [7, 10].
Note:
1 <= left <= right <= 10^9
At most 10^5 calls in total will be made to add and count.
At least one call will be made to count.
解析
根据题意，给定一组空的区间集合，实现一个数据结构，它可以：
- 将间隔添加到间隔集。
- 计算至少一个区间中存在的整数个数。
实现 CountIntervals 类：
- CountIntervals() 用一组空的间隔来初始化对象
- void add(int left, int right) 将区间 [left, right] 添加到区间集合中，这是闭区间
- int count() 返回出现在至少一个区间内的整数个数
比赛的时候我知道这道题要用到二分法，但是最后写的代码又臭又长，后来看了大佬的解释才知道，原来这是在考察珂朵莉树算法，大家可以去看一下大佬的解释，我这里就不多介绍了。
整体的思路就是在执行 add 的时候，将新来的 [left,right] 区间尽量融入到与其有覆盖交集的更大的区间中，这样就可以将区间集合不断缩小，当然如果没有覆盖的区间，那么我们就在区间集合中新增加一个区间。
时间复杂度为 O(NlogM) 因为一共调用 N 次，每次时间复杂度为 O(logM) ，M 为区间集合数量。空间复杂度为 O(M)，可能会有 M 个区间段 ，N 是调用方法的次数。
解答
from sortedcontainers import SortedDict
class CountIntervals(object):
def __init__(self):
self.result = 0
self.d = SortedDict()
def add(self, left, right):
"""
:type left: int
:type right: int
:rtype: None
"""
idx = self.d.bisect_left(left)
while idx<len(self.d) and self.d.values()[idx] <= right:
r, l = self.d.popitem(idx)
left = min(l, left)
right = max(r, right)
self.result -= r-l+1
self.result += right - left + 1
self.d[right] = left
def count(self):
"""
:rtype: int
"""
return self.result
运行结果
73 / 73 test cases passed.
Status: Accepted
Runtime: 1366 ms
Memory Usage: 54 MB
原题链接
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