一维dp最简单常用: Result[n]=E{ fun0(Result[i],w[i][n]), fun1 }, i=0~n-1
Word Break
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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, givens = "leetcode",dict = ["leet", "code"].
Return true because "leetcode" can
be segmented as "leet code".
递推方程: res[n]=E( res[i]&&word[i+1][n] , || ),
其中res......
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