Palindrome Partitioning

最新推荐文章于 2021-02-15 17:35:21 发布

转载最新推荐文章于 2021-02-15 17:35:21 发布 · 45 阅读

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原文链接：http://www.cnblogs.com/reynold-lei/p/3442591.html

本文介绍了一种使用动态规划方法解决字符串回文分隔问题的算法实现，通过构建状态转移矩阵和利用回文判断函数，有效地找到了所有可能的回文分割方案。代码示例清晰地展示了算法流程。

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"]
  ]

采用DP来做。为了不重复计算palindrome， 使用一个map 存储从[i,j]是否为palindrome， 是为1， 不是为-1，没有测试过为0.

 1 public class Solution {
 2     ArrayList<ArrayList<String>> result = null;
 3     int[][] map = null;
 4     public ArrayList<ArrayList<String>> partition(String s) {
 5         // IMPORTANT: Please reset any member data you declared, as
 6         // the same Solution instance will be reused for each test case.
 7         result = new ArrayList<ArrayList<String>>();
 8         map = new int[s.length()][s.length()]; 
 9         getPartition(s, 0, new ArrayList<String>());
10         return result;
11     }
12     public void getPartition(String s, int pos, ArrayList<String> row){
13         if(pos == s.length()) result.add(row);
14         for(int i = pos; i < s.length(); i ++){
15             if(map[pos][i] == 0){
16                 if(checkPartition(s, pos, i)) map[pos][i] = 1;
17                 else map[pos][i] = -1;
18             }
19             if(map[pos][i] == 1){
20                 row.add(s.substring(pos, i + 1));
21                 getPartition(s, i + 1, new ArrayList<String>(row));
22                 row.remove(row.size() - 1);
23             }
24         }
25     }
26     
27     public boolean checkPartition(String s, int start, int end){
28         for(int i = 0; i < (end - start + 1) / 2; i ++){
29             if(s.charAt(start + i) != s.charAt(end - i)) return false;
30         }
31         return true;
32     }
33 }

转载于:https://www.cnblogs.com/reynold-lei/p/3442591.html