【LeetCode-131】Palindrome Partitioning

这道题有些难度，一步一步调试了半天才看明白递归的过程

public class PalindromePartitioning {//生成标志回文字符串的数组，partitioning_map[i][j] = 1的话，表明：string[i..j]是一个回文字符串//如果s.charAt(i) == s.charAt(j)，partitioning_map[i + 1][j - 1] = 1，则partitioning_map[i][j] = 1//得到的这个dp[][]是一个上三角阵，下三角都是0，对角线都是1    public void dp(String s, byte[][] palindrome_map) {          for(int i = s.length() - 1;i >= 0;i --) {              for(int j = i;j < s.length();j ++) {                  if(i == j) {                      palindrome_map[i][j] = 1;                  }                 else {                      if(s.charAt(i) == s.charAt(j)) {                          if(j == i + 1 || palindrome_map[i + 1][j - 1]== 1) {                              palindrome_map[i][j] = 1;                          }                      }                  }              }          }      }          //根据生成好的回文标记数组对字符串进行划分      public void dfs(String s, int begin, byte[][] palindrome_map, ArrayList<String> array, ArrayList<ArrayList<String>> result) {          if(begin == s.length()) {              result.add(array);              return;          }                    for(int i = begin;i < s.length();i ++) {              if(palindrome_map[begin][i] == 1) {                  ArrayList<String> tmp = new ArrayList<String>(array);                  tmp.add(s.substring(begin,i + 1));                  dfs(s,i + 1,palindrome_map,tmp,result);              }          }      }          public ArrayList<ArrayList<String>> partition(String s) {          ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();          ArrayList<String> array = new ArrayList<String>();            if(s == null || s.length() == 0) {              result.add(array);              return result;          }          byte[][] palindrome_map = new byte[s.length()][s.length()];                    dp(s, palindrome_map);          dfs(s,0,palindrome_map,array,result);          return result;      }