[N-0] refactor 60

Author: fishercoder1534

README.md

@@ -601,7 +601,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |63|[Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_63.java)|O(m*n)|O(m*n)|Medium| DP
 |62|[Unique Paths](https://leetcode.com/problems/unique-paths/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_62.java)|O(m*n)|O(m*n)|Medium| DP
 |61|[Rotate List](https://leetcode.com/problems/rotate-list/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_61.java)|O(n)|O(1)|Medium| Linked List
-|60|[Permutation Sequence](https://leetcode.com/problems/permutation-sequence/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_60.java)|?|?|Medium|
+|60|[Permutation Sequence](https://leetcode.com/problems/permutation-sequence/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_60.java)|O(n^2)|O(n)|Medium| Math, Backtracking
 |59|[Spiral Matrix II](https://leetcode.com/problems/spiral-matrix-ii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_59.java)|O(n)|O(n)|Medium|
 |58|[Length of Last Word](https://leetcode.com/problems/length-of-last-word/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_58.java)|O(n)|O(1)|Easy|
 |57|[Insert Intervals](https://leetcode.com/problems/insert-interval/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_57.java)|O(n)|O(1)|Hard| Array, Sort

src/main/java/com/fishercoder/solutions/_60.java

@@ -1,6 +1,8 @@
 package com.fishercoder.solutions;
 
 /**
+ * 60. Permutation Sequence
+ *
  * The set [1,2,3,…,n] contains a total of n! unique permutations.
 
  By listing and labeling all of the permutations in order,
@@ -18,27 +20,29 @@ We get the following sequence (ie, for n = 3):
  */
 public class _60 {
 
-    public String getPermutation(int n, int k) {
-        int[] nums = new int[n + 1];
-        int permcount = 1;
-        for (int i = 0; i < n; i++) {
-            nums[i] = i + 1; // put 1, 2, 3 ... n into nums[]
-            permcount *= (i + 1);
-        }
+    public static class Solution1 {
+        public String getPermutation(int n, int k) {
+            int[] nums = new int[n + 1];
+            int permcount = 1;
+            for (int i = 0; i < n; i++) {
+                nums[i] = i + 1; // put 1, 2, 3 ... n into nums[]
+                permcount *= (i + 1);
+            }
 
-        k--;
-        StringBuilder sb = new StringBuilder();
-        for (int i = 0; i < n; i++) {
-            permcount = permcount / (n - i);
-            int idx = k / permcount;// the index that this position should
-            // choose
-            sb.append(nums[idx]);
-            // left shift nums[] by one bit
-            for (int j = idx; j < n - i; j++) {
-                nums[j] = nums[j + 1];
+            k--;
+            StringBuilder sb = new StringBuilder();
+            for (int i = 0; i < n; i++) {
+                permcount = permcount / (n - i);
+                int idx = k / permcount;// the index that this position should
+                // choose
+                sb.append(nums[idx]);
+                // left shift nums[] by one bit
+                for (int j = idx; j < n - i; j++) {
+                    nums[j] = nums[j + 1];
+                }
+                k %= permcount;
             }
-            k %= permcount;
+            return sb.toString();
         }
-        return sb.toString();
     }
 }