一个处理Range List的面试题解法

大纲

题目
解法
- Range
- - add
  - remove
- Tools
- RangeList
- - add
  - remove
代码

最近看到一个比较有意思的面试题。题目不算难，但是想把效率优化做好，也没那么容易。

我们先看下题目

题目

javascript 复制代码

// Task: Implement a class named 'RangeList'
// A pair of integers define a range, for example: [1, 5). This range includes integers: 1, 2, 3, and 4.
// A range list is an aggregate of these ranges: [1, 5), [10, 11), [100, 201)
/**
*
 * NOTE: Feel free to add any extra member variables/functions you like.
*/
class RangeList {
	/**
	*
	* Adds a range to the list
	* @param {Array<number>} range - Array of two integers that specify beginning and
	end of range.
	*/
	add(range) {
		// TODO: implement this
	}
	
	/**
	*
	* Removes a range from the list
	* @param {Array<number>} range - Array of two integers that specify beginning and
	end of range.
	*/
	remove(range) {
		// TODO: implement this
	}
	
	/**
	*
	* Convert the list of ranges in the range list to a string
	* @returns A string representation of the range list
	*/
	toString() {
		// TODO: implement this
	}
}
// Example run
const rl = new RangeList();
rl.toString(); // Should be ""
rl.add([1, 5]);
rl.toString(); // Should be: "[1, 5)"
rl.add([10, 20]);
rl.toString(); // Should be: "[1, 5) [10, 20)"
rl.add([20, 20]);
rl.toString(); // Should be: "[1, 5) [10, 20)"
rl.add([20, 21]);
rl.toString(); // Should be: "[1, 5) [10, 21)"
rl.add([2, 4]);
rl.toString(); // Should be: "[1, 5) [10, 21)"
rl.add([3, 8]);
rl.toString(); // Should be: "[1, 8) [10, 21)"
rl.remove([10, 10]);
rl.toString(); // Should be: "[1, 8) [10, 21)"
rl.remove([10, 11]);
rl.toString(); // Should be: "[1, 8) [11, 21)"
rl.remove([15, 17]);
rl.toString(); // Should be: "[1, 8) [11, 15) [17, 21)"
rl.remove([3, 19]);
rl.toString(); // Should be: "[1, 3) [19, 21)

这题大体的意思是：设计一个RangeList类，它保存了一批左闭右开的区间。它支持add操作，可以新增一个包含区间，但是可能会影响之前的区间，比如之前的区间是：[3,5) [7,9)，新增区间[5,7)之后，区间就变成[3,9)；它还支持remove操作，可以删除一个区间，也可能影响之前的区间，比如之前的区间是[3,9)，删除[5,7)之后，变成[3,5) [7,9)。

还有一种特殊区间需要考虑，就是左右值相等的区间。比如[5,5)代表的是一个空区间。

解法

Range

首先我们设计一个Range类，它只是单个区间。

add

如果对其进行add操作，即新增一个区间，则要考虑这两个区间是否相交。如果相交，则返回一个重新整合过的区间；如果不相交，则抛出异常。

python 复制代码

    # add the other range to this range.For example, [1, 5) add [5, 7) is [1, 7).
    # @param other - the other range to add to this range
    # @return - the new range after adding
    # @throws TypeError if other is not a Range object or a list of integers
    # @throws ValueError if other is not a list of 2 integers
    # @throws TypeError if other range is not overlap with this range
    def add(self, other) -> object:
        other = self.conv(other)
        
        if self.end < other.start or self.start > other.end:
            raise ValueError("other range must be overlap with this range")
        if self.start >= other.start and self.end <= other.end:
            return Range(other.start, other.end)
        if self.start >= other.start and self.end > other.end:
            return Range(other.start, self.end)
        if self.start < other.start and self.end <= other.end:
            return Range(self.start, other.end)
        if self.start < other.start and self.end > other.end:
            return Range(self.start, self.end)

remove

如果对其进行remove操作，即删除一个区间，也要考虑两个区间相交的情况。如果相交，则返回一个Range数组，其中可能0~2个区间。

python 复制代码

    # remove the other range from this range.For example, [1, 5) remove [2, 3) is [1, 2) [3, 5).
    # @param other - the other range to remove from this range.the other range must be a Range object or a list of 2 integers
    # @return - a list of Range objects that are the result of removing other from this range
    # @throws TypeError if other is not a Range object or a list of integers
    # @throws ValueError if other is not a list of 2 integers
    def remove(self, other) -> list:
        other = self.conv(other)
        
        if self.end < other.start or self.start > other.end:
            return [self]
        if self.start >= other.start and self.end <= other.end:
            return []
        if self.start >= other.start and self.end > other.end:
            return [Range(other.end, self.end)]
        if self.start < other.start and self.end <= other.end:
            return [Range(self.start, other.start)]
        if self.start < other.start and self.end > other.end:
            return [Range(self.start, other.start), Range(other.end, self.end)]

Tools

在设计完Range类后，我们还需要解决下面两个问题：

被修正的区间有哪些
需要调整位置的区间有哪些

上图中标红的表示可能要调整区间的区域。
对于没有没有需要调整的区域，则要找到临近的区域。比如上图中第一组中，[7,8)需要找到[5,6)这组区间。如果是add操作，则需要将[7,8)插入到区间数组的[5,6)后面。

python 复制代码

#!/usr/bin/env python
# -*- coding: utf-8; py-indent-offset:4 -*-
# ======================================================================================================================
# Copyrigth (C) 2024 fangliang <304646673@qq.com>
#
# This program is free software; you can redistribute it and/or modify it under the terms of the GNU Lesser General Public
# License as published by the Free Software Foundation; either version 2.1 of the License, or (at your option) any later
# version.
#
# ======================================================================================================================

from rangelist.range import Range

class Tools(object):          
        
    # search the index of the range which contains the value.First value is the index of the range where to compare with the value, 
    # second value is True if the range contains the value, False otherwise.  
    # @param ranges - the list of ranges
    # @param value - the value to search
    # @param start_index - the start index of the ranges to search
    # @return the index of the range where to compare with the value, True if the range contains the value, False otherwise
    @staticmethod
    def search(ranges, value, start_index = 0):
        if start_index < 0:
            start_index = 0
        end_index = len(ranges) - 1
        while start_index <= end_index:
            mid = (start_index + end_index) // 2
            if ranges[mid].start <= value and ranges[mid].end >= value:
                return (mid, True)
            elif ranges[mid].end < value:
                start_index = mid + 1
            else:
                end_index = mid - 1
        
        return (end_index, False)
    
    # search the index of the ranges which overlap with the search range.
    # First value is the index of the range where to compare with the value, second value is True if the range contains the value,
    # False otherwise.
    # @param ranges - the list of ranges
    # @param search_range - the range to search
    # @return a list of (index, overlap) of the ranges which overlap with the search range
    @staticmethod
    def search_overlap(ranges, search_range):
        if search_range.start == search_range.end:
            return []
        
        start = Tools.search(ranges, search_range.start)
        end = Tools.search(ranges, search_range.end, start[0])
        index_list = [start]
        for i in range(start[0]+1, end[0]+1):
            index_list.append((i, True))
        return index_list

search_overlap方法返回的数据如下：

python 复制代码

[(-1, False), (0, True), (1, True)]

-1代表对比的区间（可能是新增或者删除）的起始值在第0个区间的左侧。

True和False表示区间是否会调整（因为有覆盖）。

RangeList

RangeList用于保存一组Range序列。

这题的解法也主要依赖于其add和remove方法。

add

python 复制代码

    # add a range to the list.For example, [[1, 5)] add [2, 3) is [[1, 5)].[[1, 5)] add [6, 8) is [[1, 5) [6, 8)].
    # @param other - the other range to compare with
    # @return True if the other range is overlap with this range, False otherwise
    # @throws TypeError if other is not a Range object or a list of integers
    # @throws ValueError if other is not a list of 2 integers
    def add(self, other):
        other = Range.conv(other)
        
        indexes = Tools.search_overlap(self.ranges, other)
        del_start_index = -1
        for i in indexes:
            if i[1]:
                other = self.ranges[i[0]].add(other)
                if -1 == del_start_index:
                    del_start_index = i[0]
                    
        if -1 != del_start_index:
            del self.ranges[del_start_index : indexes[-1][0]+1]
            self.ranges.insert(del_start_index, other)
        elif len(indexes) > 0:
            self.ranges.insert(indexes[0][0]+1, other)
        
        return self

add方法会让一个Range不停"合并"被其覆盖的Range。然后删除被覆盖的Range，把新组合的Range插入到第一个覆盖的Range位置。

如果没有覆盖的区间，则在适当的位置插入。

remove

python 复制代码

# remove the other range from this range.For example, [[1, 5) [10, 14)]] remove [2, 3) is [[1, 2) [3, 5) [10, 14)]].
    # @param other - the other range to remove from this range
    # @return - the new range after removing
    # @throws TypeError if other is not a Range object or a list of integers
    # @throws ValueError if other is not a list of 2 integers
    def remove(self, other):
        other = Range.conv(other)
        
        indexes = Tools.search_overlap(self.ranges, other)
        del_start_index = -1
        range_list_from_remove_all = []
        for i in indexes:
            if i[1]:
                range_list_from_remove = self.ranges[i[0]].remove(other)
                if range_list_from_remove != None:
                    range_list_from_remove_all.extend(range_list_from_remove)
                if -1 == del_start_index:
                    del_start_index = i[0]
        
        if -1 != del_start_index:
            del self.ranges[del_start_index : indexes[-1][0]+1]
            self.ranges[del_start_index:del_start_index] = range_list_from_remove_all
        
        return self

remove方法则是让Range List中Range不停remove待删除Range，最后把切割的Range重新插入到Range List中。

代码

https://github.com/f304646673/rangelist.git