LeetCode -- Palindrome Partitioning

题目描述：


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"]
  ]


把字符串s进行分割，要求分割后的每个子串（s[i...j]，其中i,j∈[0,n)）都是回文。


本题算是回溯问题，与Combination Sum的1和2属于同类问题。


使用经典的回溯模型：
BackTracking(start , current, result):
if start is end :
add current to result


i = start to end:
currentAdd(the [i]th element);
BackTracking(i+1, current, result)
currentRemove(the [i]th element)


至于判断回文的过程就不介绍了，判断两头字符是否相等，直到中间元素位置。




实现代码：

public class Solution {    public IList<IList<string>> Partition(string s)     {        if(string.IsNullOrEmpty(s))        {    return new List<IList<string>>();    }        var result = new List<IList<string>>();    Travel(s, 0, new List<string>() , ref result);        return result;    }private void Travel(string s , int start, IList<string> current, ref List<IList<string>> result){if(start == s.Length){result.Add(new List<string>(current));return;}for(var i = start + 1; i <= s.Length; i++){var x = s.Substring(start, i - start);if(IsP(x)){current.Add(x);Travel(s, i , current, ref result);current.RemoveAt(current.Count - 1);}}}private bool IsP(string s){if(s.Length == 1){return true;}var len = s.Length % 2 == 0 ? s.Length / 2 : (s.Length - 1) / 2;for(var i = 0;i < len; i++){if(s[i] != s[s.Length - 1- i]){return false;}}return true;}}