LeetCode 题解（152）: Palindrome Partitioning

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题目：

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"]  ]

题解：

递归， Backtracing。仅当当前子串为Palindrome时，递归后半。

C++版：

class Solution {public:    vector<vector<string>> partition(string s) {        vector<vector<string>> result;        vector<string> output;        recursion(s, 0, output, result);        return result;    }        void recursion(string s, int start, vector<string>& output, vector<vector<string>>& result) {        if(start == s.length()) {            result.push_back(output);            return;        }                for(int i = start; i < s.length(); i++) {            if(isPalindrome(s, start, i)) {                output.push_back(s.substr(start, i - start + 1));                recursion(s, i + 1, output, result);                output.pop_back();            }        }    }        bool isPalindrome(string s, int start, int end) {        while(start < end) {            if(s[start++] != s[end--])                return false;        }        return true;    }};

Java版：

public class Solution {    public List<List<String>> partition(String s) {        List<List<String>> result = new ArrayList<List<String>>();        List<String> output = new ArrayList<String>();        recursion(s, 0, output, result);        return result;    }        void recursion(String s, int start, List<String> output, List<List<String>> result) {        if(start == s.length()) {            ArrayList<String> temp = new ArrayList<>();            temp.addAll(output);            result.add(temp);            return;        }                for(int i = start; i < s.length(); i++) {            if(isPalindrome(s, start, i)) {                output.add(s.substring(start, i+1));                recursion(s, i + 1, output, result);                output.remove(output.size()-1);            }        }    }        public boolean isPalindrome(String s, int start, int end) {        while(start < end) {            if(s.charAt(start++) != s.charAt(end--))                return false;        }        return true;    }}

Python版：

class Solution:    # @param {string} s    # @return {string[][]}    def partition(self, s):        result = []        output = []        self.recursion(s, 0, output, result)        return result            def recursion(self, s, start, output, result):        if start == len(s):            result.append(copy.copy(output))                    for i in range(start, len(s)):            if self.is_palindrome(s, start, i):                output.append(s[start:i+1])                self.recursion(s, i + 1, output, result)                output.pop()                    def is_palindrome(self, s, start, end):        while start < end:            if s[start] != s[end]:                return False            start += 1            end -= 1        return True

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