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
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通过题目分析最优解
- 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】
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标准求解，双层循环，鼻青脸肿但通过，超过5%
import CheckFuncPerf as cfp def merge_base(intervals): if not intervals: return [] if len(intervals) == 1: return intervals result = [] while intervals: tmpArea = intervals.pop(0) bPop = True while bPop: iPopCnt = 0 bPop = False for iIdx in range(len(intervals)): if intervals[-iIdx-1+iPopCnt][0] > tmpArea[1]: continue if intervals[-iIdx-1+iPopCnt][1] < tmpArea[0]: continue addArea = intervals.pop(-iIdx-1+iPopCnt) iPopCnt += 1 bPop = True tmpArea[0] = min(tmpArea[0], addArea[0]) tmpArea[1] = max(tmpArea[1], addArea[1]) result.append([tmpArea[0], tmpArea[1]]) return result intervals = [[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%
import CheckFuncPerf as cfp def merge_ext1(intervals): if not intervals: return [] if len(intervals) == 1: return intervals intervals.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 = newleft tmpright = newright elif newright > tmpright: tmpright = newright result.append([tmpleft, tmpright]) return result intervals = [[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 cfp
def merge_ext2(intervals):
if not intervals:
return []
if len(intervals) == 1:
return intervals
intervals.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 = newleft
tmpright = newright
else:
tmpright = max(tmpright, newright)
result.append([tmpleft, tmpright])
return result
intervals = [[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]]
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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
import CheckFuncPerf as cfp def merge_ext3(intervals): if not intervals: return [] if len(intervals) == 1: return intervals import numpy as np arr = 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 = newleft tmpright = newright elif newright > tmpright: tmpright = newright result.append([tmpleft, tmpright]) return result intervals = [[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]]一日练，一日功，一日不练十日空
may the odds be ever in your favor ~
