leetcode 字串分割對稱

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 1 public class Solution { 2     public List<List<String>> partition(String s) { 3         int len=s.length(); 4            boolean dp[][]=new boolean[len][len]; 5            get(dp,s); 6            ArrayList<ArrayList<String>> res=new ArrayList<ArrayList<String>>(); 7            ArrayList<String> temp=new ArrayList<String>(); 8            dfs(res,temp,0,dp,s); 9            return (List)res;10            11         12     }13     public void get(boolean dp[][],String s)14     {15         char c[]=s.toCharArray();16         int len=s.length();17       //single character is duichen18       for(int i=0;i<len-1;i++)19       {20           dp[i][i]=true;21           if(c[i]==c[i+1]) dp[i][i+1]=true;22           23       }24       dp[len-1][len-1]=true;25         26         for(int l=2;l<len;l++)27         {28             for(int k=0;k<len-l;k++)29             {30                 dp[k][k+l]=dp[k+1][k+l-1]&&(c[k]==c[k+l]);31             }32     33         }34         35         36         37     }38 public void dfs(ArrayList<ArrayList<String>> res,ArrayList<String> temp,int l,boolean dp[][],String s)39     {40         if(l==dp.length)41         {42            res.add(new ArrayList(temp));   43         }44         else45         {46             for(int j=0;j<dp.length;j++)47             {48                 if(dp[l][j])49                 {50                     ArrayList<String> t=new ArrayList<String>(temp);51                     t.add(s.substring(l,j+1));52                     dfs(res,t,j+1,dp,s);53                     54                 }55                     56                 57             }58                 59              }60     }61     62     63     64     65 }