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。。。