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。。。