标签：style blog color os io for art div
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"]
]
注意下面的例子：
| Input: | "fff" |
| Expected: | [["f","f","f"],["f","ff"],["ff","f"],["fff"]] |
class Solution { public: vector<vector<string>> partition(string s) { vector<vector<string>> res; if(s.size()<1) return res; vector<string> partition; addPalindrome(s,0,partition,res); return res; } private: void addPalindrome(string s,int start,vector<string> &partition,vector<vector<string> >&res){ if(start == s.size()){ res.push_back(partition); return; } for(int i = start+1;i<=s.size();i++){ string str = s.substr(start,i-start); if(isPalindrome(str)){ partition.push_back(str); addPalindrome(s,i,partition,res); partition.pop_back(); }//end if }//end for } bool isPalindrome(string str){ int left = 0; int right = str.size()-1; while(left<right){ if(str[left]!=str[right]) return false; left++; right--; }//end while return true; } };
方法：DFS！
[LeetCode] Palindrome Partitioning,布布扣,bubuko.com
[LeetCode] Palindrome Partitioning
标签：style blog color os io for art div
原文地址：http://www.cnblogs.com/Xylophone/p/3872720.html
