Expression Expand Word Break II Partition Equal Subset Sum
字符串展开问题，按照[]前的数字展开字符串，主要是维护两个栈一个是展开个数栈一个是展开内容栈，内容栈添加[用来判断是否把要展开的内容全部推出，然后要注意数字可能不止一位其他就无所谓了
class Solution:# @param {string} s an expression includes numbers, letters and brackets# @return {string} a stringdef expressionExpand(self, s):# Write your code herenl=[]
sl=[]
sc=''res=''lstr=''for i in s:if i.isdigit():if not lstr.isdigit():
sl.append(sc)
sc=''sc = sc + ielse:if i=='[':
nl.append(int(sc))
sc = ''sl.append('[')elif i==']':
n=nl.pop()while len(sl)>0:
k=sl.pop()if k== '[':breaksc = k+ sc
ts=''for j in range(n):ts= ts + sc
sc=''if len(nl) > 0:sl.append(ts)else:
res = res + tselse:if len(nl)>0:
sc = sc + ielse:
res = res + i
lstr=ireturn resWord Break II
单词切分问题，在数组重从开头的字符串开始查找，找到了就加入栈，然后每次循环pop 判断pop出的字符串是否完整，完整加入结果，不完整找后续，能找到后续加入栈，没有继续下一次循环，这里又一定歧义，wordDict里边的字符串是否可以重复使用的问题？
这玩意当字符串很长后边的字典数组很多的时候会很慢，暂时没想到什么优化的算法，lintcode给的测试数据里边好像没有这种情况只有一个特殊情况，所以前边加了一个过滤算是通过了，还有要注意python的startwith函数
如果参数是''总是返回True略坑爹····
class Solution:# @param {string} s a string# @param {set[str]} wordDict a set of wordsdef wordBreak(self, s, wordDict):# Write your code herehead=[]
ss=''for i in s:if ss=='':
ss=ielse:if i not in ss:
ss = ss + ifor i in ss:
flag=Falsefor di in wordDict:if i in di:
flag=Truebreak;if not flag:return []for di in wordDict:if di !='' and s.startswith(di):
head.append(di)if len(head)<1:return []
cur=s
res=[]while len(head)>0:
h=head.pop()
le=len(h.replace(' ',''))
cur=s[le:]if cur == '':
res.append(h)continuefor di in wordDict:if cur.startswith(di):
head.append(h+' '+di) return res
class Solution:# @param {int[]} nums a non-empty array only positive integers# @return {boolean} return true if can partition or falsedef canPartition(self, nums):# Write your code heresum=0for n in nums:
sum = sum +nif sum%2!=0:return False
k=sum//2a=[None]*len(nums)for i in range(len(nums)):
a[i]=[0]*k for i in range(len(nums)):for j in range(k):if i == 0:
a[i][j] = nums[i] if nums[i] < j+1 else 0else:if nums[i]> j+1:
a[i][j]=a[i-1][j]else:
a[i][j]=max(nums[i]+a[i-1][j+1-nums[i]],a[i-1][j]) if a[i][j] ==k:return Truereturn False
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