LeetCode笔记：Weekly Contest 327
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0. 做题感受
这次的题目一看大佬们的用时，最快的居然也花了将近30min，把我吓了一大跳，结果真正做下来就有点无语了，就是超级繁琐的各种分类讨论以及边界条件，难度倒是感觉还好，但是真要再做一遍估计还得去掉半条命，简直了……
1. 题目一
给出题目一的试题链接如下：
1. 解题思路
这一题有序数列的话其实可以二分找边界，不过我这边就比较暴力的直接遍历统计了一遍就是了……
2. 代码实现
给出python代码实现如下：
class Solution:
def maximumCount(self, nums: List[int]) -> int:
pos, neg = 0, 0
for x in nums:
if x < 0:
neg += 1
elif x > 0:
pos += 1
return max(pos, neg)
提交代码评测得到：耗时135ms，占用内存14.2MB。
3. 题目二
给出题目二的试题链接如下：
1. 解题思路
这一题其实就是不断地去最大元素进行操作，然后更新数组就行了。
问题就在于如何快速地更新数组以及取最大值，我是采用了堆排，不过全有序排序也问题不大就是了。
2. 代码实现
给出python代码实现如下：
class Solution:
def maxKelements(self, nums: List[int], k: int) -> int:
nums = [-x for x in nums]
score = 0
heapq.heapify(nums)
for _ in range(k):
x = -heapq.heappop(nums)
score += x
heapq.heappush(nums, - ceil(x/3))
return score
提交代码评测得到：耗时1132ms，占用内存29.1MB。
2. 题目三
给出题目三的试题链接如下：
1. 解题思路
这一题我这边的做法就是一个非常暴力的分类讨论，首先统计两个单词之中出现了那些字符，然后分别考察交换两个字符之后可能出现的情况，将所有情况分别枚举一边就行了……
2. 代码实现
给出python代码实现如下：
class Solution:
def isItPossible(self, word1: str, word2: str) -> bool:
cnt1 = Counter(word1)
cnt2 = Counter(word2)
diff = len(cnt1) - len(cnt2)
for ch1 in cnt1:
for ch2 in cnt2:
if ch1 == ch2:
change = 0
elif cnt1[ch1] == 1 and cnt2[ch2] == 1:
if cnt2[ch1] > 0 and cnt1[ch2] > 0:
change = 0
elif cnt2[ch1] > 0 and cnt1[ch2] == 0:
change = 1
elif cnt2[ch1] == 0 and cnt1[ch2] > 0:
change = -1
else:
change = 0
elif cnt1[ch1] == 1 and cnt2[ch2] > 1:
if cnt2[ch1] > 0 and cnt1[ch2] > 0:
change = -1
elif cnt2[ch1] > 0 and cnt1[ch2] == 0:
change = 0
elif cnt2[ch1] == 0 and cnt1[ch2] > 0:
change = -2
else:
change = -1
elif cnt1[ch1] > 1 and cnt2[ch2] == 1:
if cnt2[ch1] > 0 and cnt1[ch2] > 0:
change = 1
elif cnt2[ch1] > 0 and cnt1[ch2] == 0:
change = 2
elif cnt2[ch1] == 0 and cnt1[ch2] > 0:
change = 0
else:
change = 1
else:
if cnt2[ch1] > 0 and cnt1[ch2] > 0:
change = 0
elif cnt2[ch1] > 0 and cnt1[ch2] == 0:
change = 1
elif cnt2[ch1] == 0 and cnt1[ch2] > 0:
change = -1
else:
change = 0
if change == -diff:
return True
return False
提交代码评测得到：耗时114ms，占用内存15.4MB。
4. 题目四
给出题目四的试题链接如下：
1. 解题思路
这一题其实思路不怎么复杂，就是按照题意完全用代码翻译一下，然后模拟一下其完整的过程就可以了。
不过，如前所述，问题就是繁琐……
2. 代码实现
给出python代码实现如下：
class Solution:
def findCrossingTime(self, n: int, k: int, time: List[List[int]]) -> int:
eff = sorted([(t[0] + t[2], i) for i, t in enumerate(time)], reverse=True)
orders = [0 for _ in range(k)]
for i in range(k):
orders[eff[i][1]] = i
right, left = 0, 1
q = [(left, orders[i], i) for i in range(k)]
heapq.heapify(q)
waited = []
t = 0
while q or waited:
while q == [] and waited:
nxt, side, order, wid = heapq.heappop(waited)
if n > 0 or side == right:
t = max(t, nxt)
heapq.heappush(q, (side, order, wid))
while waited and waited[0][0] <= t:
_, side, order, wid = heapq.heappop(waited)
heapq.heappush(q, (side, order, wid))
if q == []:
break
side, order, wid = heapq.heappop(q)
if side == left:
if n == 0:
continue
t += time[wid][0]
heapq.heappush(waited, (t + time[wid][1], right, order, wid))
n -= 1
else:
t += time[wid][2]
heapq.heappush(waited, (t + time[wid][3], left, order, wid))
return t
提交代码评测得到：744ms，占用内存21.2MB。
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