新年快乐！2021年第一题，每日一题！希望2021年LC和github可以全绿！加油！
https://leetcode-cn.com/problems/can-place-flowers/
代码如下：
vllbc02
class Solution:
def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
flowerbed = [0] + flowerbed + [0]
for i in range(1,len(flowerbed)-1):
if flowerbed[i-1] == 0 and flowerbed[i] == 0 and flowerbed[i+1] == 0:
n -= 1
flowerbed[i] = 1
return n <= 0思路很暴力，就是三个0在一起就可以插进去。。
https://leetcode-cn.com/problems/coin-change/
以我目前的水平做出来有点吃力，看了思路才做出来
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [float('inf')] * (amount + 1)
dp[0] = 0
for i in range(amount+1):
for coin in coins: #
if i >= coin:
dp[i] = min(dp[i],dp[i-coin]+1)
return -1 if (dp[-1] == float("inf")) else dp[-1]伪代码如下
# 伪码框架
def coinChange(coins: List[int], amount: int):
# 定义：要凑出金额 n，至少要 dp(n) 个硬币
def dp(n):
# 做选择，选择需要硬币最少的那个结果
for coin in coins:
res = min(res, 1 + dp(n - coin))
return res
# 题目要求的最终结果是 dp(amount)
return dp(amount)一开始看到题目感觉挺简单的，没想到对现在的我挺有难度。。
https://leetcode-cn.com/problems/remove-duplicate-letters/
#1
class Solution:
def removeDuplicateLetters(s: str):
res = ""
while s: #用递归也可以
loc = min(map(s.rindex,s)) #s.rindex是返回列表各值最后出现的索引 求这个最小的索引
a = min(s[:loc+1]) #求字典序最小的
res += a
s = s[s.index(a):].replace(a,"") #把已经加入的和与其重复的都去掉了
return res
#2
#遍历字符串，压入栈，如果遇到比栈顶小的元素且当前字符后面还有与栈顶相同的元素时，移除栈顶元素
class Solution:
def removeDuplicateLetters(s: str) -> str:
stack = []
for i, t in enumerate(s):
if t in stack:
continue
while stack !=[] and t < stack[-1] and s[i:].find(stack[-1]) != -1:
stack.pop()
stack.append(t)
return "".join(stack)两个方法，第二个方法更好想点。第一个方法是copy的
Table file.mtime as "最后修改时间"
WHERE tags[0] = "Reading" and date(today) - file.mtime <= dur(7 days)
Sort file.mtime descTable rows.file.link as filename
WHERE todo and title != "<% tp.file.title %>"
Sort file.ctime desc
GROUP BY tags[1] as "category"Table file.mtime as "最后修改时间"
WHERE !todo and date(today) - file.mtime <= dur(7 days) and tags[0] != "Reading" and title
Sort file.mtime descTable file.mtime as "最后修改时间"
WHERE tags[0] = "Reading" and this.file.ctime - file.mtime <= dur(7 days)
Sort file.mtime descPLAN-AND-ACT：Improving Planning of Agents for Long-Horizon Tasks 这篇论文解答了我心中一直存在的一个想法，就是计划和行动是分离的。在agent的执行过程中，要先制定计划，然后按照计划一步步执行，计划并不是可以执行的动作，因此需要一个转换。即分离规划器和执行器。
目前想通过对 verl 项目做贡献来做为以后的项目经历，可以贡献的地方：