[leetcode] Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given

s =

"leetcode"

,

dict =

["leet", "code"]

.

Return true because

"leetcode"

can be segmented as

"leet
code"

.

第一种方法：递归（超时）Time Limit Exceeded

思路：从s的第一个字母向后匹配，如果i前面的前缀可以匹配，就看s字符串i以后的后缀是否匹配

Last executed input:

"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]

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bool wordBreak(string s, unordered_set<string> &dict) {

// Note: The Solution object is instantiated only once.

if(s.length() < 1) return true;

bool flag = false;

for(int i = 1; i <= s.length(); i++)

{

string tmpstr = s.substr(0,i);

unordered_set<string>::iterator it = dict.find(tmpstr);

if(it != dict.end())

{

if(tmpstr.length() == s.length())return true;

flag = wordBreak(s.substr(i),dict);

}

if(flag)return true;

}

return false;

}

第二种方法：dpAccepted

思路：从s的第一个字母向后匹配，如果i前面的前缀可以匹配，就看s字符串i以后的后缀是否匹配，在找后缀是否匹配时添加了记忆功能。

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bool wordBreakHelper(string s, unordered_set<string> &dict,set<string> &unmatch) {

if(s.length() < 1) return true;

bool flag = false;

for(int i = 1; i <= s.length(); i++)

{

string prefixstr = s.substr(0,i);

unordered_set<string>::iterator it = dict.find(prefixstr);

if(it != dict.end())

{

string suffixstr = s.substr(i);

set<string>::iterator its = unmatch.find(suffixstr);

if(its != unmatch.end())continue;

else{

flag = wordBreakHelper(suffixstr,dict,unmatch);

if(flag) return true;

else unmatch.insert(suffixstr);

}

}

}

return false;

}

bool wordBreak(string s, unordered_set<string> &dict) {

// Note: The Solution object is instantiated only once.

int len = s.length();

if(len < 1) return true;

set<string> unmatch;

return wordBreakHelper(s,dict,unmatch);

}

dp改进：dict中的单词有的长有的短，当prefixstr串小于最短串时就不匹配了，当prefixstr串大于最长的串时也不用匹配了。多谢@阿桂爱清净

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bool wordBreakHelper(string s,unordered_set<string> &dict,set<string> &unmatched,int mn,int mx) {

if(s.size() < 1) return true;

int i = mx < s.length() ? mx : s.length();

for(; i >= mn ; i--)

{

string preffixstr = s.substr(0,i);

if(dict.find(preffixstr) != dict.end()){

string suffixstr = s.substr(i);

if(unmatched.find(suffixstr) != unmatched.end())

continue;

else

if(wordBreakHelper(suffixstr, dict, unmatched,mn,mx))

return true;

else

unmatched.insert(suffixstr);

}

}

return false;

}

bool wordBreak(string s, unordered_set<string> &dict) {

// Note: The Solution object is instantiated only once.

if(s.length() < 1) return true;

if(dict.empty()) return false;

unordered_set<string>::iterator it = dict.begin();

int maxlen=(*it).length(), minlen=(*it).length();

for(it++; it != dict.end(); it++)

if((*it).length() > maxlen)

maxlen = (*it).length();

else if((*it).length() < minlen)

minlen = (*it).length();

set<string> unmatched;

return wordBreakHelper(s,dict,unmatched,minlen,maxlen);

}