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，如果当前的后缀没有匹配就把它放进set中，以后就不用再看这个后缀时候匹配了。
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);
}
题目刚放出来就A过了，好激动啊。。。
最近做题bug free的次数好多啊。。。