139. Word Break
2017-12-24 15:19
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139. Word Break
题目Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words. For example, given s = "leetcode", dict = ["leet", "code"]. Return true because "leetcode" can be segmented as "leet code". UPDATE (2017/1/4): The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.
解析
class Solution { public: bool wordBreak(string s, vector<string>& wordDict) { set<string> strset(wordDict.begin(), wordDict.end());//vector-->set if (wordDict.size() == 0) { return false; } //动态规划 vector<bool> dp(s.size() + 1, false); dp[0] = true; for (int i = 1; i <= s.size(); i++) //第一层遍历,s的每个位置是否可分成字典元素 { for (int j = i - 1; j >= 0; j--) { if (dp[j]) //j之前的元素可以分成字典元素 { if (strset.find(s.substr(j, i - j)) != strset.end()) { dp[i] = true; break; } } } } return dp[s.size()]; } };
题目来源
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