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  • 139. Word Break

    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();
        }
    };



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