Word Break

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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".

//使用DP算法！

//注：f(i) = f(j) && find(s.substr(j , i-j)); // f(i)表示s[0..i]可以分词！ ---理解即可一个for 针对i,一个针对j!!

class Solution {public:    //注：f(i) = f(j) && find(s.substr(j , i-j)); // f(i)表示s[0..i]可以分词！    bool wordBreak(string s, unordered_set<string> &dict) {                int length = s.size();        bool table[length+1];                if(s.empty() || dict.empty())return false;                memset(table,false,sizeof(bool)*(length+1));                table[0] = true;         for(int i = 1; i <= length ; i++)        {             for(int j = 0; j < i; j++)             {                 //substr(j , i-j) 此处的j为长度的意思，而下标从0开始，故而不用加1！                if(table[j] && dict.find(s.substr(j , i-j)) != dict.end())                {                    table[i] = true ;                }             }        }               return table[length];    }    };

增加输出最多可以划分的个数,时间复杂度O（N^2*m^2*size)，也可采取划分，再在trie树中找，时间复杂度O(N^2*m+size*m)//查找速度快！（在搜索中trie树。）

#include <iostream> #include <vector>#include <cstdlib>  #include <algorithm>#include <string>using namespace std; //注：f(i) = f(j) && find(s.substr(j , i-j)); // f(i)表示s[0..i]可以分词！  int wordBreak(string s, vector<string> &dict) {  int length = s.size();  int *table = new int[length+1];  if(s.empty() || dict.empty())return false;  memset(table,0,sizeof(int)*(length+1));  table[0] = 1;   for(int i = 1; i <= length ; i++)  {  for(int j = 0; j < i; j++)  {  //substr(j , i-j) 此处的j为长度的意思，而下标从0开始，故而不用加1！  if(table[j]>0 && find(dict.begin(),dict.end(),s.substr(j , i-j)) != dict.end())  {  table[i] = table[j]+1;  }  }  }  return table[length] > 1?table[length]-1:0;//一开始多加了1}     int main(){string str[8] = {"a", "an","ave","na","ve","bar", "nave", "on"};vector<string> dict(str,str+8);string my_str("anavebaron");int res;if(res = wordBreak(my_str,dict))cout<<"find:"<<res<<endl;elsecout<<" not find"<<endl;return 0;}

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