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"]

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。

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

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);
}