solve problem Beautiful Arrangement

Author: P-ppc

README.md

@@ -264,6 +264,7 @@ All solutions will be accepted!
 |59|[Spiral Matrix II](https://leetcode-cn.com/problems/spiral-matrix-ii/description/)|[java/py/js](./algorithms/SpiralMatrixII)|Medium|
 |885|[Spiral Matrix III](https://leetcode-cn.com/problems/spiral-matrix-iii/description/)|[java/py/js](./algorithms/SpiralMatrixIII)|Medium|
 |19|[Remove Nth Node From End Of List](https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/description/)|[java/py/js](./algorithms/RemoveNthNodeFromEndOfList)|Medium|
+|526|[Beautiful Arrangement](https://leetcode-cn.com/problems/beautiful-arrangement/description/)|[java/py/js](./algorithms/BeautifulArrangement)|Medium|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/BeautifulArrangement/README.md

@@ -0,0 +1,2 @@
+# Beautiful Arrangement
+We can solve this problem by Backtracking Algorithm

algorithms/BeautifulArrangement/Solution.java

@@ -0,0 +1,28 @@
+class Solution {
+    public int countArrangement(int N) {
+        int[] res = new int[]{0};
+        List<Integer> candidates = new ArrayList<Integer>();
+        for (int i = 1; i <= N; i++) {
+            candidates.add(i);
+        }
+        backTrack(candidates, 0, res);
+        return res[0];
+    }
+    
+    public void backTrack(List<Integer> candidates, int i, int[] res) {
+        int size = candidates.size();
+        if (size == 0) {
+            res[0]++;
+        } else {
+            i++;
+            for (int j = 0; j < size; j++) {
+                int c = candidates.get(j);
+                if (c % i == 0 || i % c == 0) {
+                    List<Integer> tempCandidates = new ArrayList<Integer>(candidates);
+                    tempCandidates.remove(j);
+                    backTrack(tempCandidates, i, res);
+                }
+            }
+        }
+    }
+}

algorithms/BeautifulArrangement/solution.js

@@ -0,0 +1,31 @@
+/**
+ * @param {number} N
+ * @return {number}
+ */
+var countArrangement = function(N) {
+    let res = [],
+        candidates = []
+    
+    for (let i = 1; i <= N; i++)
+        candidates.push(i)
+    
+    backTrack(candidates, [], res)
+    return res.length
+};
+
+var backTrack = function (candidates, temp, res) {
+    if (candidates.length == 0) {
+        res.push(temp)
+    } else {
+        let i = temp.length + 1
+        candidates.forEach((c, index) => {
+            if (c % i == 0 || i % c == 0) {
+                let tempCandidates = candidates.slice(),
+                    t = temp.slice()
+                tempCandidates.splice(index, 1)
+                t.push(c)
+                backTrack(tempCandidates, t, res)
+            }
+        })
+    }
+};

algorithms/BeautifulArrangement/solution.py

@@ -0,0 +1,22 @@
+class Solution(object):
+    def countArrangement(self, N):
+        """
+        :type N: int
+        :rtype: int
+        """
+        res = []
+        self.backtrack(range(1, N + 1), [], res)
+        return len(res)
+    
+    def backtrack(self, candidates, temp, res):
+        if len(candidates) == 0:
+            res.append(temp)
+        else:
+            i = len(temp) + 1
+            for c in candidates:
+                if c % i == 0 or i % c == 0:
+                    temp_candidates = candidates[:]
+                    temp_candidates.remove(c)
+                    t = temp[:]
+                    t.append(c)
+                    self.backtrack(temp_candidates, t, res)