Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals.
For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be:
[1, 1] [1, 1], [3, 3] [1, 1], [3, 3], [7, 7] [1, 3], [7, 7] [1, 3], [6, 7]
Follow up:
What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?
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这题二分法很容易想到,问题是找到插入位置后还要进行O(n)的移动,这点不确定,看了一下别人的题解,好像这个移动都做了,所以就跟着往下做了。。。
这题能一次提交通过主要感谢自己在 https://blog.youkuaiyun.com/taoqick/article/details/100026169 的总结,二分退出来,二分的范围是[0,l]闭区间,有了这点,分别讨论四种情况:和左边在一起,和右边在一起,merge,插入,最终codes是,遗憾的是自己在Pycharm里居然写了俩bug,不应该啊:
class SummaryRanges:
def __init__(self):
"""
Initialize your data structure here.
"""
self.segs = []
def addNum(self, val):
l = len(self.segs)
left, right = 0, l - 1
while (left <= right):
mid = left + ((right - left) >> 1) #bug1: mid = left + ((right - left) >> 1)
if (self.segs[mid][0] <= val): #bug2: self.segs[mid][0] >= val
left = mid + 1
else:
right = mid - 1
# merge
if (left - 1 >= 0 and left < l and self.segs[left - 1][1] + 1 == self.segs[left][0] - 1 and self.segs[left - 1][1] + 1 == val):
self.segs[left - 1][1] = self.segs[left][1]
self.segs.pop(left)
elif (left - 1 >= 0 and self.segs[left - 1][1] + 1 == val): #放左边
self.segs[left - 1][1] = val
elif (left < l and self.segs[left][0] - 1 == val): #放右边
self.segs[left][0] = val
elif ((left - 1 >= 0 and self.segs[left - 1][1] + 1 < val) or left == 0): #必须插入
self.segs.insert(left, [val, val])
def getIntervals(self):
return self.segs
# Your SummaryRanges object will be instantiated and called as such:
obj = SummaryRanges()
obj.addNum(1)
print(obj.getIntervals())
obj.addNum(3)
print(obj.getIntervals())
obj.addNum(7)
print(obj.getIntervals())
obj.addNum(2)
print(obj.getIntervals())
obj.addNum(6)
print(obj.getIntervals())