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

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- 问题分析

1. 本题为求区间数组的合并
2. 主要的计算为2个，1区间遍历，2区间合成

- 优化思路

1. 减少循环层次

2. 增加分支，减少计算集合

3. 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

4. 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】

---

1. 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]]
   

2. 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]]
   

3. 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]]

4. 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 ~