finish "131. Palindrome Partitioning". Fix #131

Author: diguage

Diff for: README.adoc

@@ -658,12 +658,12 @@
 //|{leetcode_base_url}/surrounded-regions/[Surrounded Regions]
 //|{source_base_url}/SurroundedRegions.java[Java]
 //|Medium
-//
-//|131
-//|{leetcode_base_url}/palindrome-partitioning/[Palindrome Partitioning]
-//|{source_base_url}/PalindromePartitioning.java[Java]
-//|Medium
-//
+
+|131
+|{leetcode_base_url}/palindrome-partitioning/[Palindrome Partitioning]
+|{source_base_url}/PalindromePartitioning.java[Java]
+|Medium
+
 //|132
 //|{leetcode_base_url}/palindrome-partitioning-ii/[Palindrome Partitioning II]
 //|{source_base_url}/PalindromePartitioningIi.java[Java]

Diff for: src/main/java/com/diguage/algorithm/leetcode/PalindromePartitioning.java

@@ -0,0 +1,74 @@
+package com.diguage.algorithm.leetcode;
+
+import java.util.*;
+
+/**
+ * = 131. Palindrome Partitioning
+ *
+ * https://leetcode.com/problems/palindrome-partitioning/[Palindrome Partitioning - LeetCode]
+ *
+ * Given a string s, partition s such that every substring of the partition is a palindrome.
+ *
+ * Return all possible palindrome partitioning of s.
+ *
+ * .Example:
+ * [source]
+ * ----
+ * Input: "aab"
+ * Output:
+ * [
+ *   ["aa","b"],
+ *   ["a","a","b"]
+ * ]
+ * ----
+ *
+ * @author D瓜哥, https://www.diguage.com/
+ * @since 2020-01-04 19:04
+ */
+public class PalindromePartitioning {
+    /**
+     * Runtime: 2 ms, faster than 94.23% of Java online submissions for Palindrome Partitioning.
+     * Memory Usage: 39.2 MB, less than 95.45% of Java online submissions for Palindrome Partitioning.
+     *
+     * Copy from: https://leetcode.com/problems/palindrome-partitioning/discuss/41963/Java%3A-Backtracking-solution.[Java: Backtracking solution. - LeetCode Discuss]
+     */
+    public List<List<String>> partition(String s) {
+        if (Objects.isNull(s) || s.length() == 0) {
+            return Collections.emptyList();
+        }
+        List<List<String>> result = new LinkedList<>();
+        List<String> current = new ArrayList<>();
+        dfs(s, 0, current, result);
+        return result;
+    }
+
+    private void dfs(String s, int index, List<String> current, List<List<String>> result) {
+        if (index == s.length()) {
+            result.add(new ArrayList<>(current));
+        } else {
+            for (int i = index; i < s.length(); i++) {
+                if (isPalindrome(s, index, i)) {
+                    current.add(s.substring(index, i + 1));
+                    dfs(s, i + 1, current, result);
+                    current.remove(current.size() - 1);
+                }
+            }
+        }
+    }
+
+    private boolean isPalindrome(String s, int low, int high) {
+        while (low < high) {
+            if (!Objects.equals(s.charAt(low++), s.charAt(high--))) {
+                return false;
+            }
+        }
+        return true;
+    }
+
+    public static void main(String[] args) {
+        PalindromePartitioning solution = new PalindromePartitioning();
+        String s1 = "aab";
+        List<List<String>> r1 = solution.partition(s1);
+        System.out.println(Arrays.deepToString(r1.toArray()));
+    }
+}