760 Find Anagram Mappings

You are given two integer arrays nums1 and nums2 where nums2 is an anagram of nums1. Both arrays may contain duplicates.

Return an index mapping array mapping from nums1 to nums2 where mapping[i] = j means the ith element in nums1 appears in nums2 at index j. If there are multiple answers, return any of them.

An array a is an anagram of an array b means b is made by randomizing the order of the elements in a.

Example 1:

Input: nums1 = [12,28,46,32,50], nums2 = [50,12,32,46,28]
Output: [1,4,3,2,0]
Explanation:
 As mapping[0] = 1 because the 0th element of nums1 appears at nums2[1], and mapping[1] = 4 because the 1st element of nums1 appears at nums2[4], and so on.

Example 2:

Input: nums1 = [84,46], nums2 = [84,46]
Output: [0,1]

Constraints:

• 1 <= nums1.length <= 100

• nums2.length == nums1.length

• 0 <= nums1[i], nums2[i] <= 105

• nums2 is an anagram of nums1.

还是一如既往的简单题，查java doc比想算法的时间还长。因为有dup，所以map的位置用了list来存。花了S:O(N),T:O(N)

public int[] anagramMappings(int[] nums1, int[] nums2) {
    if (nums1 == null || nums2 == null || nums1.length != nums2.length) {
        return null;
    }
    
    int n = nums1.length;
    int[] result = new int[n];
    
    // <val, loc list>
    Map<Integer, List<Integer>> locMap = new HashMap<>();
    for (int i = 0; i < n; i++) {
        if (locMap.containsKey(nums2[i])) {
            locMap.get(nums2[i]).add(i);
        } else {
            List<Integer> tmp = new LinkedList<>();
            tmp.add(i);
            locMap.put(nums2[i], tmp);
        }
    }
    
    for (int i = 0; i < n; i++) {
        List<Integer> tmp = locMap.get(nums1[i]);
        result[i] = tmp.remove(0);
    }
    return result;
}
}j