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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, given
s = "leetcode"
,
dict = ["leet", "code"]
.
Return true because "leetcode"
can be segmented as "leet code"
.
[Solution]
对于S=“leetcode”, 我们用table[1...n]记录S[0...n-1]是否为dictionary word.
下面可以用动态规划求解。
假设已知table[1...k], 即已知S[0, ..., k-1]是否为dictionary word,那么
如果table[1] = true, 只要S[1, ..., k]是dictionary word, 则table[k] = true
如果table[2] = true, 只要S[2, ..., k]是dictionary word, 则table[k] = true
...
可得table[k + 1] = (FOR ANY i IN [0, k], IF table[i] = true && S[i, ..., k] IS DICT WORD)
1 bool wordBreak(string s, unordered_set<string> &dict) 2 { 3 vector<bool> table(s.size() + 1, false); 4 table[0] = true; 5 6 for (int i = 1; i <= s.size(); i++) 7 { 8 for (int j = 0; j < i; j++) 9 { 10 if (table[j] == false) 11 continue; 12 if (dict.find(s.substr(j, i - j)) != dict.end()) 13 { 14 table[i] = true; 15 break; 16 } 17 } 18 } 19 20 return table[s.size()]; 21 }
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原文地址:http://www.cnblogs.com/ym65536/p/4227361.html