139. Word Break

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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.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false



这题很自然的暴力解法是遍历字符串,每个子串去匹配dict里面的word,一直到结束,如果都能匹配上就返回true
优化方案: 把dict做成哈希表会更快一点,题目默认给的是vector

class Solution {
public:
    bool wordBreak(string s, vector<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])continue;
                string word=s.substr(j,i-j);
                for(int k=0;k<wordDict.size();++k)
                {
                    if(wordDict[k]!=word)continue;
                    dp[i]=true;
                    break;
                }
                if(dp[i])
                    break;
            }
        return dp.back();
    }
};

139. Word Break

标签：Plan   nbsp   ted   cti   time   out   ret   string   rdb   

原文地址：https://www.cnblogs.com/lychnis/p/12034487.html