LeetCode Palindrome Partitioning

链接: https://oj.leetcode.com/problems/palindrome-partitioning/

dp+dfs

求字符串s的所有分割,使得每一个子串都是回文字符串.

dp数组用来储存s[i]~s[j]是否为回文串.由于二维数组传递不方便,所以用dp[i]来计算  :dp[i][j] ->dp[i*n+j]

if s[j]==s[j+i]&& dp[j+1][j+i-1]==true   ;->dp[j][j+i]=true;

dfs(i,...) :求s[i~s.length()]的所有符合条件的分割.

class Solution
{
public:

    void dfs(int i,string s,int n,bool dp[] ,vector<string> &vet,vector<vector<string> > &ans)
    {

        if(i==n)
        {
            ans.push_back(vet);
            return ;
        }

        for(int j=i; j<n; j++)
        {
            if(dp[i*n+j])
            {
                vet.push_back(s.substr(i,j-i+1));
                dfs(j+1,s,n,dp,vet,ans);
                vet.pop_back();

            }
        }
    }
    vector<vector<string> >partition(string s)
    {
        int n=s.size();
        bool *dp=new bool[n*n];


        vector<vector<string> > ans;
        vector<string> vet;
        if(n==0)
            return ans;

        for(int i = 0; i < n*n; i ++)
            dp[i] = false;
        for(int i = 0; i < n; i++)
        {
            dp[i*n+i] = true;			
        }
        for(int i = 1; i < n; i ++)
        {
            for(int j = 0; j+i < n; j ++)
            {
                if(s[j] == s[j+i] && (dp[(j+1)*n+j+i-1]||i==1))
                    dp[j*n+j+i] = true;
            }
        }
        dfs(0,s,n,dp,vet,ans);
        return ans;
    }
};