Min swaps to sort array

Given an array with distinct numbers, return an integer indicating the minimum number of swap operations required to sort the array into ascending order.

Example 1:

Input: [5, 1, 3, 2]
Output: 2
Explanation: [5, 1, 3, 2] -> [2, 1, 3, 5] -> [1, 2, 3, 5]

Example 2:

Input: [1, 3, 2]
Output: 1
Explanation: [1, 3, 2] -> [1, 2, 3]
分析：
先把原数组保存一份，然后对另一份排序，并且把原数组的值与index保存一份。这样，我们需要得到某个数的位置的时候，直接通过map得到。

 public class Solution {
     public int minSwaps(int[] elems) {
         int counter = ;
         Map<Integer, Integer> map = buildMap(elems);
         int[] copy = new int[elems.length];
         System.arraycopy(elems, , copy, , elems.length);
         Arrays.sort(elems);

         for (int i = ; i < copy.length; i++) {
             if (copy[i] != elems[i]) {
                 swap(copy, i, map.get(elems[i]), map);
                 counter++;
             }
         }
         return counter;
     }

     private Map<Integer, Integer> buildMap(int[] elems) {
         Map<Integer, Integer> map = new HashMap<Integer, Integer>();
         for (int i = ; i < elems.length; i++) {
             map.put(elems[i], i);
         }
         return map;
     }

     private void swap(int[] elems, int i, int j, Map<Integer, Integer> map) {
         map.put(elems[j], i);
         map.put(elems[i], j);
         int temp = elems[j];
         elems[j] = elems[i];
         elems[i] = temp;
     }
 }