[Leetcode]131.palindrome Partitioning

Topic

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

Return all possible palindrome partitioning of S.

For example, given s = "AaB",
Return

[
["AA", "B"],
["A", "a", "B"]
]

Ideas

This problem can be solved by backtracking. Divide the string s into front and back two strings str1, str2, if STR1 is a palindrome, add partition, and then recursively str2.

Code

    /**------------------------------------* Date: 2015-03-02 * SJF0115 * Title: 131.Palindrome Partitioning * URL: https://oj.leetcode.com/problems/palindrome-partitioning/* result: AC * Source: Leetcode * Blog:----------- ----------------------------**/    #include <iostream>    #include <vector>    #include <algorithm>    using namespace STD;classSolution { Public: vector<vector<string> >Partitionstrings) { vector<string>Path vector<vector<string> >ResultintSize = S.size ();if(Size <=0){returnResult }//ifPartition (S,size,0, Path,result);returnResult }Private://s source string size source string length start split point        //Path Intermediate results result end result        voidPartition (stringStrintSizeintStart vector<string>&path, vector<vector<string> >&result) {//Termination conditions            if(start = = size) {result.push_back (path);return; }//if            stringSubstr//Split string             for(inti = Start;i < Size;++i) {substr = Str.substr (start,i-start+1);//Determine if it is a palindrome string                if(Ispalindrome (SUBSTR))                    {Path.push_back (substr); Partition (str,size,i+1, Path,result);                Path.pop_back (); }//if}//for}//Determine if the string is a palindrome        BOOLIspalindrome (stringSTR) {intSize = Str.size ();if(Size = =0){return false; }//if            intleft =0;intright = size-1; while(Left < right) {if(Str[left]! = Str[right]) {return false; }//ifleft++;            right--; }//while            return true; }    };intMain () {solution S;stringStr"Aaba"); vector<vector<string> >result = S.partition (str);//Output         for(inti =0; i < result.size (); ++i) { for(intj =0; J < Result[i].size (); ++j) {cout<<result[i][j]<<" "; }//for            cout<<endl; }//for        return 0; }

Run time

[Leetcode]131.palindrome Partitioning