Python算法题集_合并区间-编程知识

Python算法题集_合并区间

本文为Python算法题集之一的代码示例

题目56：合并区间

说明：以数组 intervals 表示若干个区间的集合，其中单个区间为 intervals[i] = [starti, endi] 。请你合并所有重叠的区间，并返回 一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间 。

示例 1：

输入：intervals = [[1,3],[2,6],[8,10],[15,18]]
输出：[[1,6],[8,10],[15,18]]
解释：区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].

示例 2：

输入：intervals = [[1,4],[4,5]]
输出：[[1,5]]
解释：区间 [1,4] 和 [4,5] 可被视为重叠区间。

提示：

1 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104

- 问题分析

本题为求区间数组的合并
主要的计算为2个，1区间遍历，2区间合成

- 优化思路

减少循环层次
增加分支，减少计算集合
通过题目分析最优解
1. 区间数组排序后，运算情况会大为简单
采用内置算法提升计算效率

CheckFuncPerf是我写的函数用时和内存占用模块，地址在这里：Python算法题集_检测函数用时和内存占用的模块【自搓】

标准求解，双层循环，鼻青脸肿但通过，超过5% 在这里插入图片描述

import CheckFuncPerf as cfpdef merge_base(intervals):if not intervals:return []if len(intervals) == 1:return intervalsresult = []while intervals:tmpArea = intervals.pop(0)bPop = Truewhile bPop:iPopCnt = 0bPop = Falsefor iIdx in range(len(intervals)):if intervals[-iIdx-1+iPopCnt][0] > tmpArea[1]:continueif intervals[-iIdx-1+iPopCnt][1] < tmpArea[0]:continueaddArea = intervals.pop(-iIdx-1+iPopCnt)iPopCnt += 1bPop = TruetmpArea[0] = min(tmpArea[0], addArea[0])tmpArea[1] = max(tmpArea[1], addArea[1])result.append([tmpArea[0], tmpArea[1]])return resultintervals = [[1, 3], [2, 6], [8, 10], [15, 18]]
result = cfp.getTimeMemoryStr(merge_base, intervals)
print(result['msg'], '执行结果 = {}'.format(result['result']))# 运行结果
函数 merge_base 的运行时间为 1.00 ms；内存使用量为 4.00 KB 执行结果 = [[1, 6], [8, 10], [15, 18]]

标准求解，区间数组排序，单层循环，超过85% 在这里插入图片描述

import CheckFuncPerf as cfpdef merge_ext1(intervals):if not intervals:return []if len(intervals) == 1:return intervalsintervals.sort(key=lambda item: (item[0], item[1]))result = []tmpleft, tmpright = intervals[0][0], intervals[0][1]for index in range(1, len(intervals)):newleft, newright = intervals[index]if newleft > tmpright:result.append([tmpleft, tmpright])tmpleft = newlefttmpright = newrightelif newright > tmpright:tmpright = newrightresult.append([tmpleft, tmpright])return resultintervals = [[1, 3], [2, 6], [8, 10], [15, 18]]
result = cfp.getTimeMemoryStr(merge_ext1, intervals)
print(result['msg'], '执行结果 = {}'.format(result['result']))# 运行结果
函数 merge_ext1 的运行时间为 0.00 ms；内存使用量为 0.00 KB 执行结果 = [[1, 6], [8, 10], [15, 18]]

标准求解，单层循环，神之一刀，超越97%

比较再赋值改为单次max计算

import CheckFuncPerf as cfpdef merge_ext2(intervals):if not intervals:return []if len(intervals) == 1:return intervalsintervals.sort(key=lambda item: (item[0]))result = []tmpleft, tmpright = intervals[0][0], intervals[0][1]for index in range(1, len(intervals)):newleft, newright = intervals[index]if newleft > tmpright:result.append([tmpleft, tmpright])tmpleft = newlefttmpright = newrightelse:tmpright = max(tmpright, newright)result.append([tmpleft, tmpright])return resultintervals = [[1, 3], [2, 6], [8, 10], [15, 18]]
result = cfp.getTimeMemoryStr(merge_ext2, intervals)
print(result['msg'], '执行结果 = {}'.format(result['result']))# 运行结果
函数 merge_ext2 的运行时间为 1.00 ms；内存使用量为 4.00 KB 执行结果 = [[1, 6], [8, 10], [15, 18]]

探索第三方排序，失败 采用numpy排序，不作不死，超过6%，内存也爆 在这里插入图片描述

import CheckFuncPerf as cfpdef merge_ext3(intervals):if not intervals:return []if len(intervals) == 1:return intervalsimport numpy as nparr = np.array(intervals)tmplist = np.sort(arr, 0).tolist()result = []tmpleft, tmpright = tmplist[0][0], tmplist[0][1]for index in range(1, len(tmplist)):newleft, newright = tmplist[index]if newleft > tmpright:result.append([tmpleft, tmpright])tmpleft = newlefttmpright = newrightelif newright > tmpright:tmpright = newrightresult.append([tmpleft, tmpright])return resultintervals = [[1, 3], [2, 6], [8, 10], [15, 18]]
result = cfp.getTimeMemoryStr(merge_ext3, intervals)
print(result['msg'], '执行结果 = {}'.format(result['result']))# 运行结果
函数 merge_ext3 的运行时间为 116.01 ms；内存使用量为 14944.00 KB 执行结果 = [[1, 6], [8, 10], [15, 18]]