palindrome-partitioning

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Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s =”aab”,
Return

[
[“aa”,”b”],
[“a”,”a”,”b”]
]

class Solution {
public:
 //判断是否为回文数
    bool palindrome(char *p,int len){
    char*q=p+len-1;

    while(q>p){
        if(*p==*q){
            q--;
            p++;
        }else{
            return false;
        }
    }

    return true;
}   
//深度优先搜索解空间
    void dfs(char* p,string s,int i,int j,vector< vector<string> >&result,vector<string>temp){

    //  vector<string> temp;

            //  vector<string> temp;
    //for(int k=j;k>=0;k--)
        for(int l=i;l<=j;l++){

            //如果i 至 l 是回文数 
            if(palindrome(p+i,l-i+1)){
                temp.push_back(s.substr(i,l-i+1));

                if(l==j){
                result.push_back(temp);
                }else{
                    dfs(p,s,l+1,j,result,temp); 
                }   
                temp.pop_back();
            }

            //如果此段不是回文数 循环继续； 
        }

}
    vector<vector<string>> partition(string s) {
        int len=s.length();
        char *p=new char[len+1];
        for(int i=0;i<len;i++)
            p[i]=s[i];
        p[len]='\0';

        vector<string>temp;
        vector< vector<string> > result;

        dfs(p,s,0,len-1,result,temp);

        return result;
    }
};