[Leetcode]Palindrome Partitioning

dyllanzhou

于 2015-10-24 15:03:39 发布

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分类专栏： Backtracking 文章标签： leetcode 算法

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本文介绍了如何使用深度优先搜索（DFS）算法解决字符串回文分组问题，即找到一个字符串的所有可能的回文子串分组。

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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"]
  ]

Have you met this question in a real interview?

class Solution {
public:
    /*algorithm: DFS
    
    */
    bool isPalindrome(string &s){//"aab"
        for(int i = 0,j = (int)s.size() - 1;i < j;i++,j--){
            if(s[i] != s[j])return false;
        }
        return true;
    }
    void dfs(vector<vector<string>>&result,vector<string>&path,string s){
        if(s.empty()){
            result.push_back(path);
            return;
        }
        for(int i = 1;i <= s.size();i++){
            string sub = s.substr(0,i);
            if(!isPalindrome(sub))continue;
            path.push_back(sub);
            dfs(result,path,s.substr(i));
            path.pop_back();
        }
    }
    vector<vector<string>> partition(string s) {
                vector<vector<string>>result;
                vector<string>path;
                dfs(result,path,s);
                return result;
    }
};