leetcode wordBreak

最新推荐文章于 2022-02-13 19:30:15 发布

cch13

最新推荐文章于 2022-02-13 19:30:15 发布

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分类专栏： leetcode 题目

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本文介绍了一种判断字符串是否能通过字典中的单词进行分割的方法，并提供了三种不同的实现方式：使用备忘录的动态规划、自底向上的动态规划及字典树方法。这些算法有助于理解和解决字符串分割问题。

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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".

#include <string>
#include <unordered_set>
#include <iostream>
using namespace std;
class Node{
public:
    bool end;
    Node *node[26];
    Node() : end(false)
    {
        for (int i = 0; i < 26; i++)
            node[i] = NULL;
    }
    ~Node()
    {
        for (int i = 0; i < 26; i++)
            delete node[i];
    }
    
    void append(string s)
    {
        Node *iter = this;
        int len = s.length();

        for(int i = 0; i < len; i++)
        {
            if(iter->node[s[i] - 'a'] == NULL)
               iter->node[s[i] - 'a'] = new Node();
            iter = iter->node[s[i] - 'a'];
        }
        iter->end = true;
    }
};
class Solution {
private:
    int **f;
    int n;
public:
    Solution(): f(NULL){}
    
    ~Solution()
    {
        if(f != NULL)
        {
            for(int i = 0; i < n; i++)
                delete []f[i];
            delete []f;

        }
    }

    //动态规划之备忘录做法
    bool wordBreakI(string s, unordered_set<string> &dict) 
    {
        n = s.length();
        f = new int*[n];
        for(int i = 0; i < n; i++)
            f[i] = new int[n];
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                  f[i][j] = 0;
        
        return wordBreak(s, dict, 0, n);
    }
    bool wordBreak(string s, unordered_set<string> &dict, int start, int end)
    {
        if (f[start][end - 1] == -1)
            return false;
        if (f[start][end - 1] == 1)
            return true;

        string subs = s.substr(start, end - start);

        if (dict.count(subs) != 0)  //如果存在从start开始的长度为end - start的string
        {
            f[start][end - 1] = 1;
            return true;
        }
        else
        {
            f[start][end - 1] = -1;
        }

        for (int j = start + 1; j < end; j++)
        {
            if(wordBreak(s, dict, start, j) && wordBreak(s, dict, j, end))
            {
                f[start][end - 1] = 1;
                return true;
            }
        }

        return false;    
    }

    //动态规划之自底向上求法
    bool wordBreakII(string s, unordered_set<string> &dict) 
    {
       vector<bool> b(s.length() + 1, false);
       b[0] = true;

       int len = s.length() + 1;
       

       //判断substr(0, i)能不能被分解
       for (int i = 1; i < len; i++)
           for (int j = i - 1; j >= 0; j--)
           {
               if (b[j] && dict.count(s.substr(j, i - j)) != 0)
                   b[i] = true;
           }
       return b[s.length()];
    }
    
    //字典树方法

    bool wordBreak(string s, unordered_set<string> &dict) 
    {
        Node node;
        unordered_set<string>::iterator end = dict.end();
        for (unordered_set<string>::iterator it = dict.begin(); it != end; it++)
            node.append(*it);
        int len = s.length();

        vector<bool> b(s.length(), false);
        
        findMatchWord(s, &node, 0, b);

        for (int i = 0; i < len; i++)
        {
            if (b[i])
                findMatchWord(s, &node, i + 1, b);
        }

        return b[s.length() - 1];
    }

    void findMatchWord(string &s, Node *iter, int start, vector<bool> &b)
    {
        int i = start;
        int len = s.length();
        while (i < len)
        {
            if (iter->node[s[i] - 'a'] != NULL)
            {
                if (iter->node[s[i] - 'a']->end)
                    b[i] = true;

                iter = iter->node[s[i] - 'a'];
            }
            else
            {
                break;
            }
            i++;
        }
    }
    

};

int main()
{
	Solution s;
    string ss("ab");
    unordered_set<string> unset;
    unset.insert(string("a"));
    unset.insert(string("b"));
    cout << s.wordBreak(ss, unset);
    return 0;
}