135.Candy
# Share
# There are N children standing in a line. Each child is assigned a rating value.
#
# You are giving candies to these children subjected to the following requirements:
#
# Each child must have at least one candy.
# Children with a higher rating get more candies than their neighbors.
# What is the minimum candies you must give?
from math import *
class Solution:
def candy(self, ratings):
n = len(ratings)
tmp = [1] * n
for i in range(n-1):
if ratings[i] < ratings[i+1]:
tmp[i+1] = tmp[i]+1
# print(tmp)
for i in range(n-1,0,-1):
if ratings[i-1] > ratings[i]:
tmp[i-1] = max(tmp[i]+1,tmp[i-1])
# print(tmp)
return sum(tmp)
if __name__=='__main__':
a = Solution()
res = [1,2,3,4,2,1]
print(a.candy(res))
139. Word Break
# Given a non-empty string s and a dictionary wordDict containing a list of non-empty words,
# determine if s can be segmented into a space-separated sequence of one or more dictionary words.
#
# Note:
#
# The same word in the dictionary may be reused multiple times in the segmentation.
# You may assume the dictionary does not contain duplicate words.
# solve:
# dp[j]=true if dp[i-1]=true && s[i:j+1] in dict
from math import *
from typing import *
class Solution:
def wordBreak(self, s, wordDict):
dp = [False]*len(s)
for i in range(len(s)):
for j in range(i, len(s)):
if s[i:j+1] in wordDict and (dp[i-1]== True or i-1 == -1):
dp[j] = True
print(dp)
return dp[-1]
if __name__=='__main__':
a = Solution()
s = "leetcode"
wordlist = ["leet", "code"]
print(a.wordBreak(s,wordlist))
140. Word Break II
# Given a non-empty string s and a dictionary wordDict containing a list of non-empty words,
# add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
#
# Note:
#
# The same word in the dictionary may be reused multiple times in the segmentation.
# You may assume the dictionary does not contain duplicate words.
# 這題不能算水題吧，畢竟自己的做法僅比6%的人快，權當python複習吧
# 下面提供兩種做法，第一種就是利用139的結論，但是效率很低，第二種是用字典存當前str能夠分解的種類
# 代碼通俗易懂，看下就行
from math import *
from typing import *
class Solution:
def judge(self,s,dict):
dp = [False] * (len(s)+1)
dp[0] = True
for i in range(1,len(s)+1):
for j in range(0,i):
if dp[j] and s[j:i] in dict:
dp[i] = True
return dp[len(s)]
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
ans=[]
tmp=""
if s in wordDict:
ans.append(s)
for i in range(len(s)):
tmp+=s[i]
if tmp in wordDict:
if self.judge(s[i+1:],wordDict):
res = self.wordBreak(s[i+1:],wordDict)
for j in res:
ans.append(tmp+" "+j)
return ans
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
return self.helper(s, wordDict, {})
def helper(self, s, wordDict, res):
if s in res:
return res[s]
if not s:
return []
ans = []
for word in wordDict:
if not s.startswith(word):
continue
if len(word) == len(s):
ans.append(word)
else:
t = self.helper(s[len(word):], wordDict, res)
for item in t:
item = word + ' ' + item
ans.append(item)
res[s] = ans
# print(res)
return ans
if __name__=='__main__':
a = Solution()
# TEST
s = "catsanddog"
lis = ["cat", "cats", "and", "sand", "dog"]
print(a.wordBreak(s,lis))