我大概算了一下，一天查60个单词，坚持十天也才600个单词，一个月也就1800个，但是如果坚持一年，那就是上万了。所以说，很多事情都是一点一点做起来的。
Wuthering heights
convulsive 痉挛的
prettier pretty
pint 一品脱 473.17毫升
endeavor 努力 effort，struggle
snatch 抢，grasp，hook，seize
bade bid 投标
echo 回声 repeat，duplicate
utmost 极度的 extreme
constitution 章程 foundation
wager 担保 stake，bet
lull calm 使镇静
bench 长凳，工作台
asunder 分开地
contradict 违背 oppose，deny
purse 追求 chase，seek
consent or denial 同意 或者 拒绝
venturesome 冒险的 adventure
pledge 担保，典当 guarantee
retract 撤回 cancel，recall
foremost 首先 first
injudicious judicial 明断的
clasp hands trembled 握紧，grasp，hook，颤抖 vibrate，shake
superstitious superstition 迷信
sup sip
accent 口音
indignation 愤怒
beggar 乞丐
contradiction paradox
ferret seek，look for
agitation 骚动
heedless careless
bolt nuts and bolts 螺帽和螺栓
audibly audible 听得见的
vociferate utter loudly
clamorously clamorous noisy
sunbeam 光束
piercing chill，very cold
haggard 憔悴的
prevail popular
extinguish kill，destroy，annihilate
ember 余烬
eavesdropper 偷听者
inarticulate 嘴笨的
lavish generous
delirium extreme excitement
whey 乳清
saucier saucy，rude，cheeky
passionate passion 热烈的
阳光刚好透过窗户找到我的键盘上
这一次有几段话，我觉得很有意思
"why do you love him, Miss Cathy"
"Nonsense, I do -that is sufficient "
"By no means, you must say why?"
"Well, because he is handsome, and pleasant to be with."
"Bad!" was my commentary
"And because he is young and cheerful."
"Bad, still."
"And because he loves me."
"Indifferent, coming there."
"And he will be rich, and I shall like to be the greatest woman of the neighbourhood, and I shall be proud of having such a husband.
Worst of all, And now, say how you love him?"
"As everybody loves - You are silly, Nelly"
"Not at all. - Answer"
"I love the ground under his feet, and the air over his head, and everything he touches, and every word he says. I love all his looks, and all his actions, and him entirely and altogether. There now."
"And why?"
"Nay, - you are making a jest of it, it is exceedingly ill-natured! It is no jest to me!" said the yound lady, scowling, and turning her face to the fire.
"I am very far from jesting, Miss Catherine". I replied. "You love Mr. Edgar because he is handsome, and young, and cheerful, and rich, and loves you. The last(划重点，他爱你不是你爱他的理由，太深刻了！), however, goes for nothing: you would love him without that, probably; and with it you wouldn't, unless he possessed the four former attractions."
"No, to be sure not: I should only pity him (这就是添狗的下场)- hate him, perhaps, if he were ugly, and a clown."
But there are several other handsome, rich young men in the world: handsomer, possibly, and richer than he is. What should hinder you from loving him?"
"If there be any, they are out of my way! I've seen none like Edgar."
"You may see some; and he won't always be handsome, and young, and may not always be rich."
"He is now; and I have only to do with the prersent. I wish you would speak rationally."
"Well, that settles it; If you have only to do with the present, marry Mr. Linton."
抄到这里差不多了，勃朗特已经说了很多重点了，真正的爱不只是皮囊，爱一个人只会让人卑微到尘埃里（张爱玲说的很对），最后，时间不等人。
现在来揭晓谜底，凯瑟琳到底爱谁，为什么爱他
对象
Heathcliff
理由如下
because he is more myself than I am. Whatever our souls are made of, his and mine are the same; and Linton's is as different as a moonbeam from ligntning, or froest from fire.（爱，是可以改变一个人的）
找到原因，凯瑟琳是在寻找她的灵魂伴侣。也就是说，不光男性会有白月光和蚊子血，女性也是一样贪婪。
杠精如我，忍不住想问凯瑟琳一个问题，
Absolutely right， but if you find another soulmate in this university, will you chase your lightning, your fire and leave Heathcliff?
好了，肚子饿了，吃午饭去了
代码
题目挺简单的，主要用了一个特殊的break语句，结束本次循环
class Solution:
def findContentChildren(self, g: List[int], s: List[int]) -> int:
num = 0
s.sort()
for biscuit in s:
for child in g:
if child<=biscuit:
num+=1
g.remove(child)
break
return num
时间有点长了
class Solution:
def canCross(self, stones: List[int]) -> bool:
re = False
step = 1
for i in range(len(stones)-1):
lenth = stones[i+1]-stones[i]
if lenth > step+1:
break
else:
step = lenth
if i == len(stones)-2:
re = True
return re
不考虑隔着石子跳过河的情况，通过率居然80%，也是醉了
又是用递归解决的，本质上是一种遍历搜索，真心不希望遇见这样的考题，我会直接选择放弃，这几行实在是太烧脑了。
# If there are no matchsticks, then we can't form any square
nums = matchsticks
if not nums:
return False
# Number of matchsticks we have
L = len(nums)
# Perimeter of our square (if one can be formed)
perimeter = sum(nums)
# Possible side of our square.
possible_side = perimeter // 4
# If the perimeter can be equally split into 4 parts (and hence 4 sides, then we move on).
if possible_side * 4 != perimeter:
return False
# Reverse sort the matchsticks because we want to consider the biggest one first.
nums.sort(reverse=True)
# This array represents the 4 sides and their current lengths
sums = [0 for _ in range(4)]
# Our recursive dfs function.
def dfs(index):
# If we reach the end of matchsticks array, we check if the square was formed or not
if index == L:
# If 3 equal sides were formed, 4th will be the same as these three and answer should be True in that case.
return sums[0] == sums[1] == sums[2] == possible_side
# The current matchstick can belong to any of the 4 sides (provided their remaining lenghts are >= the size of the current matchstick)
for i in range(4):
# If this matchstick can fir in the space left for the current side
if sums[i] + nums[index] <= possible_side:
# Recurse
sums[i] += nums[index]
if dfs(index + 1):
return True
# Revert the effects of recursion because we no longer need them for other recursions.
sums[i] -= nums[index]
return False
return dfs(0)
头大，这种题目真的可以跳过了
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