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题目：
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".
这题拿到就开始硬做。。是不对滴。用DP做怎样？搞清楚state，function，initialize和result。
1 public boolean wordBreak(String s, Set<String> dict) { 2 int maxWordLen = 0; 3 for (String str : dict) { 4 maxWordLen = Math.max(maxWordLen, str.length()); 5 } 6 //F[i] 代表在i时，前面的是否可以被字典中的词拆分 7 //F[i] = (F[j] && dict.contains(s.substring(j ,i))) 8 //F[0] = true; 9 //return F[s.length()] 10 boolean[] F = new boolean[s.length() + 1]; 11 F[0] = true; 12 for (int i = 1; i <= s.length(); i++) { 13 F[i] = false; 14 for (int j = 0; j < i && j <= maxWordLen; j++) { 15 if (F[i - j - 1] && dict.contains(s.substring(i - j - 1, i))){ 16 F[i] = true; 17 } 18 } 19 } 20 return F[s.length()]; 21 }
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原文地址：http://www.cnblogs.com/gonuts/p/4413402.html
