Merge pull request #2368 from keepgogogo/master

回溯算法篇 9 分割回文串，Java题解示例代码提供使用动态规划优化回文串判断的实现代码

problems/0131.分割回文串.md

@@ -349,6 +349,65 @@ class Solution {
 }
 ```
 
+### Java
+回溯+动态规划优化回文串判断
+```Java
+class Solution {
+    List<List<String>> result;
+    LinkedList<String> path;
+    boolean[][] dp;
+
+    public List<List<String>> partition(String s) {
+        result = new ArrayList<>();
+        char[] str = s.toCharArray();
+        path = new LinkedList<>();
+        dp = new boolean[str.length + 1][str.length + 1];
+        isPalindrome(str);
+        backtracking(s, 0);
+        return result;
+    }
+
+    public void backtracking(String str, int startIndex) {
+        if (startIndex >= str.length()) {
+            //如果起始位置大于s的大小，说明找到了一组分割方案
+            result.add(new ArrayList<>(path));
+        } else {
+            for (int i = startIndex; i < str.length(); ++i) {
+                if (dp[startIndex][i]) {
+                    //是回文子串，进入下一步递归
+                    //先将当前子串保存入path
+                    path.addLast(str.substring(startIndex, i + 1));
+                    //起始位置后移，保证不重复
+                    backtracking(str, i + 1);
+                    path.pollLast();
+                } else {
+                    //不是回文子串，跳过
+                    continue;
+                }
+            }
+        }
+    }
+
+    //通过动态规划判断是否是回文串,参考动态规划篇 52 回文子串
+    public void isPalindrome(char[] str) {
+        for (int i = 0; i <= str.length; ++i) {
+            dp[i][i] = true;
+        }
+        for (int i = 1; i < str.length; ++i) {
+            for (int j = i; j >= 0; --j) {
+                if (str[j] == str[i]) {
+                    if (i - j <= 1) {
+                        dp[j][i] = true;
+                    } else if (dp[j + 1][i - 1]) {
+                        dp[j][i] = true;
+                    }
+                }
+            }
+        }
+    }
+}
+```
+
 ### Python 
 回溯 基本版
 ```python