[Leetcode click to take notes] palindrome partitioning

Given a stringS, PartitionSSuch that every substring of the partition is a palindrome.

Return all possible palindrome partitioningS.

For example, givenS="aab",
Return

  [    ["aa","b"],    ["a","a","b"]  ]

Question: Use Aab to describe the algorithm:

1. Check whether the first character is a and whether it is a background. If it is a background, recursively judge whether the remaining AB is a background;
2. Check the first two characters AA to determine whether to return the text. If the result is a return text, recursively judge whether the remaining B is returned;
3. Check the first three characters of AAB to determine whether it is a text return. If it is not a text return.
4. The loop ends.

So this typical recursive algorithm: first judge whether s (0, I) is a return. If so, save S (0, I) to the result list, then recursively judge S (I + 1, S. length) indicates whether it is a return. When returning recursively, take S (0, I) out of the result and continue to judge whether s (0, I + 1) is a return ...... in the recursive process, if the s passed to the recursive function is an empty string, it indicates that a segmentation method is found and stored in the result list, then, the result is placed in the final answer list answer.

The Code is as follows:

 1 public class Solution { 2     private boolean isPar(String s){ 3         int begin = 0; 4         int end = s.length() - 1; 5          6         while(begin < end){ 7             if(s.charAt(begin) != s.charAt(end)) 8                 return false; 9             10             begin++;11             end--;12         }13         14         return true;15     }16     public void partitionDfs(String s,List<String> result,List<List<String>> answer){17         if(s.length() == 0){18             List<String> temp = new ArrayList<String>(result);19             answer.add(temp);20             return;21         }22         23         int length = s.length();24         for(int i = 1;i <= length;i++){25             String sub = s.substring(0,i);26             if(isPar(sub)){27                 result.add(sub);28                 partitionDfs(s.substring(i), result, answer);29                 result.remove(result.size()-1);30             }31         }32         33     }34     public List<List<String>> partition(String s) {35         List<List<String>> answer = new ArrayList<List<String>>();36         List<String> result = new ArrayList<String>();37         partitionDfs(s, result, answer);38         39         return answer;40     }41 }