Leetcode 352. Data Stream as Disjoint Intervals

最新推荐文章于 2021-10-09 02:53:45 发布

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本文介绍了一种算法，用于处理数据流输入的非负整数，将这些数字汇总为不相交的区间列表。通过使用二分查找法确定插入位置，并在不同情况下（如与左侧或右侧合并、单独插入等）更新区间，该算法能够有效地处理大量合并操作，即使当不相交区间的数量远小于数据流大小时也能保持高效。

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())