Word Break

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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”.

呃……DP？

感觉代码写的挺不好的……

AC代码：

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class Solution {
public:
	vector<vector<int>> DP;
	bool solve(int begin, int end, string& s, unordered_set<string>& wordDict)
	{
		if (begin > end) return false;
		if (DP[begin][end] == 1) return true;
		else if (DP[begin][end] == 2) return false;
		if (begin == end)
		{
			string str;
			str.push_back(s[begin]);
			if (wordDict.find(str) != wordDict.end())
			{
				DP[begin][end] = 1; return true;
			}
			else
			{
				DP[begin][end] = 2; return false;
			}
		}
		else if (begin > end)
		{
			DP[begin][end] = 2;
			return false;
		}
		else
		{
			string str;
			for (int i = begin; i <= end; ++i)
			{
				str.push_back(s[i]);
			}
			if (wordDict.find(str) != wordDict.end())
			{
				DP[begin][end] = 1; return true;
			}

			for (int i = begin; i <= end; ++i)
			{
				bool l = solve(begin, i, s, wordDict);
				bool r = solve(i+1, end, s, wordDict);
				if (l && r)
				{
					DP[begin][end] = 1; return true;
				}
				else
				{
					if (0 == DP[begin][end])
					{ DP[begin][end] = 2; }
				}
			}
		}
		DP[begin][end] = 2;
		return false;
	}

	bool wordBreak(string s, unordered_set<string>& wordDict) {
		DP = vector<vector<int>>(s.length(),vector<int>(s.length(),0));
		return solve(0, s.length()-1, s, wordDict);
	}
};