题目: https://leetcode.com/problems/word-break-ii/
难度:
Medium
还是backtracking，会超时
class Solution(object): # 此法超时
def wordBreak(self, s, wordDict):
"""
:type s: str
:type wordDict: Set[str]
:rtype: bool
"""
self.res = []
self.wordBreakLst("",s,wordDict)
return self.res
def wordBreakLst(self, lst, rest, wordDict):
if rest == '':
self.res.append(lst.rstrip())
# print lst
for i in range(1+len(rest)):
if rest[:i] in wordDict:
self.wordBreakLst(lst + rest[:i] + " ",rest[i:],wordDict)
然后看到有把word break i 结合起来减少时间复杂度的作法。
做法如下，聪明：
就是对于每一个s，我们来check它是否可以break，如果不可以，就不用做相应的操作了
class Solution(object):
def wordBreak(self, s, wordDict):
"""
:type s: str
:type wordDict: List[str]
:rtype: List[str]
"""
self.res = []
self.wordBreakLst(s, wordDict, '')
return self.res
def check(self, s, wordDict):
ok = [True]
for i in range(1, len(s) + 1):
ok += any(ok[j] and s[j:i] in wordDict for j in range(i)),
return ok[-1]
def wordBreakLst(self, s, wordDict, stringLst):
if self.check(s, wordDict):
if len(s) == 0 : self.res.append(stringLst[1:]) # 因为最开始也加了一个空格
for i in range(1,len(s)+1):
if s[:i] in wordDict:
self.wordBreakLst(s[i:], wordDict, stringLst + ' ' + s[:i])但是其实
s = "aaaaaa"
wordDict = ["a","aa","aaa"]
print a.wordBreak(s,wordDict)还是会loop很多次
不过像
s = "aabbb"
wordDict = ["a","abbb"]
就会极其的减少loop次数
看看stefan大神的做法：
sentences(i) returns a list of all sentences that can be built from the suffix s[i:].
class Solution(object):
def wordBreak(self, s, wordDict):
"""
:type s: str
:type wordDict: List[str]
:rtype: List[str]
"""
memo = {len(s): ['']}
def sentences(i):
if i not in memo:
memo[i] = [s[i:j] + (tail and ' ' + tail)
for j in range(i+1, len(s)+1)
if s[i:j] in wordDict
for tail in sentences(j)]
return memo[i]
return sentences(0)