[LeetCode] Word Break

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".

分析:动态规划D[i] = D[j] && substring[j, i] 在字典内 这个表达式对于j = 0- (i-1)任何一个成立

class Solution {
public:
    bool wordBreak(string s, unordered_set<string>& wordDict) {
        vector<bool> dp(s.size()+1, false);
        dp[0] = true;
        
        for (int i = 1; i <= s.size(); i++) {
            for (int j = i - 1; j >= 0; j--) {
                if (dp[j]) {
                    string word = s.substr(j, i-j);
                    if (wordDict.find(word) != wordDict.end()) {
                        dp[i] = true;
                        break;
                    }
                }
            }
        }
        
        return dp[s.size()];
    }
};

参考资料:http://www.senonweb.com/?p=480

原文地址:https://www.cnblogs.com/vincently/p/4817659.html