Word Break II

最新推荐文章于 2017-07-17 19:50:03 发布

原创最新推荐文章于 2017-07-17 19:50:03 发布 · 1k 阅读

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本文介绍了一种算法，该算法能够将一个字符串通过查找字典中的有效词汇进行拆分，并返回所有可能的句子组合。示例中使用了catsanddog作为输入字符串，通过查找包含cat cats and sand dog的字典来构造可能的句子。

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Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

class Solution {
public:
    void printRes(int n,int* matrix,string s,vector<string>& ret,int j,stack<string> st)
    {
        if(j == -1){
            string str = "";
            while(!st.empty()){
                str+= st.top()+" ";
                st.pop();
            } 
            if(str!="") str.erase(str.size()-1,1);
            ret.push_back(str);
            return;
        }
        for(int i = 0;i<n;i++)
        {
            if(matrix[i*n+j] == 2){
                st.push(s.substr(i,j-i+1));
                printRes(n,matrix,s,ret,i-1,st);
                st.pop();
            }
        }
        
    }
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<string> ret;
        int n = s.length();
        int matrix[n][n];
        for(int i = 0;i<n;i++)
            memset(matrix[i],0,n*sizeof(int));
        
        for(int l = 1;l<=n;l++)
        {
            for(int i = 0;i<n-l+1;i++)
            {
                int j = i+l-1;
              
                if(dict.find(s.substr(i,l)) != dict.end()) {
                    matrix[i][j] = 2;
                    continue;
                }
                for(int k = i;k<j;k++)
                {
                    if(matrix[i][k] != 0 && matrix[k+1][j] != 0) matrix[i][j] = 1;
                }
            }
        }
       
        stack<string> st;
        printRes(n,(int *)matrix,s,ret,n-1,st);
        return ret;   
    }
};

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