LeetCode OJ 之 Word Break (断词)
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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"
.
思路:
动态规划法:
设状态为f ( i ),表示前i个字符s[0,i-1] 是否可以分词,则状态转移方程为
f ( i ) = any_of (f (j )&&s [j + 1, i ] ∈ dict ), 0 ≤ j <i
代码:
class Solution {public: bool wordBreak(string s, unordered_set<string> &dict) { //f ( i ) = any_of (f (j )&&s [j + 1, i ] ∈ dict ), 0 ≤ j < i vector<bool> f(s.size()+1,false); f[0] = true; for(int i = 1 ; i <= s.size() ; i++) { for(int j = 0 ; j < i ; j++) { f[i] = f[j] && (dict.find(s.substr(j,i-j)) != dict.end()); if(f[i]) break; } } return f[s.size()]; }};
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