twitter oa 一問

gavin5566 发表于 2016-10-26 14:42
本題

判斷ipv4 or ipv6

能再解释一下 ipv6这题 具体是什么啊??

gavin5566 发表于 2016-10-27 04:30
ex
[1,2,2,3]; k = 4

array里面的元素 有没有说全是正数??   还是正负 都有可能??  leetcode有个类似的 是求最短subarray sum >= k正好跟这个相反, 但是限制了array里面的数全是正数

skysbjdy 发表于 2016-10-27 08:03
能再解释一下 ipv6这题 具体是什么啊??

判斷string 是ipv4 or ipv6

也是遇到aggregation这题，也是一个case都不过，好怀疑人生，怀疑他们定的答案有问题。。。

array a[], k 那道題有人做完了嗎? 我test case 9,10過不了

# Enter your code here. Read input from STDIN. Print output to STDOUT
# need to check if in range
import sys
import collections

def validTimeRange (curTime, timeRange):
    minYear = int(timeRange[0][:4])
    minMonth = int(timeRange[0][-2:])
    maxYear = int(timeRange[1][:4])
    maxMonth = int(timeRange[1][-2:])
    curYear = int(curTime[:4])
    curMonth  = int(curTime[-2:])
    if (curYear < minYear or curYear > maxYear or
        (curYear == minYear and curMonth < minMonth) or
        (curYear == maxYear and curMonth >= maxMonth)):
        return False
    else:
        return True
   
   
def splitWithOptionalSpace (curLine, nextLine):
    engagement_index = 11
    if (nextLine[11] == ' '):
        engagement_index += 1
    engagement = ''
    while(nextLine[engagement_index] != ','):
        engagement += nextLine[engagement_index]
        engagement_index += 1
    curLine.append(engagement)
    count_index = engagement_index + 1
    if (nextLine[count_index] == ' '):
        count_index += 1
    count = ''
    while(nextLine[count_index] != '\n'):
        count += nextLine[count_index]
        count_index += 1
    curLine.append(count)
   
   
   
timeRange = sys.stdin.readline().split(', ')

sys.stdin.readline()

nextLine = sys.stdin.readline()
# string : {}
# {} is string : int
ret = {}
while (nextLine != ''):
    curLine = [nextLine[:10]]
    # deal with fields seperated by optional space
    splitWithOptionalSpace (curLine, nextLine)
    # test if time is in valid range
    curTime = curLine[0][:7]
    if (validTimeRange(curTime, timeRange) is False):
        nextLine = sys.stdin.readline()
        continue
    # aggregate one line to the structure we want
    if curTime not in ret:
        ret[curTime] = {curLine[1] : int(curLine[2])}
    else:
        if curLine[1] in ret[curTime]:
            ret[curTime][curLine[1]] += int(curLine[2])
        else:
            ret[curTime][curLine[1]] = int(curLine[2])
   
    nextLine = sys.stdin.readline()

# format output
for entry in sorted(ret.items(), key=lambda x: x[0], reverse=True):
    toPrint = entry[0] + ', '
    for subEntry in sorted(entry[1].items(), key=lambda x: x[0]):#, reverse=True):
        toPrint += subEntry[0] + ', ' + str(subEntry[1]) + ', '
    sys.stdout.write(toPrint[:-2] + '\n')
            

全过

同问这道题，楼主解决了吗～