Formatted question description: https://leetcode.ca/all/760.html
760. Find Anagram Mappings (Easy)
Given two lists A
and B
, and B
is an anagram of A
. B
is an anagram of A
means B
is made by randomizing the order of the elements in A
.
We want to find an index mapping P
, from A
to B
. A mapping P[i] = j
means the i
th element in A
appears in B
at index j
.
These lists A
and B
may contain duplicates. If there are multiple answers, output any of them.
For example, given
A = [12, 28, 46, 32, 50] B = [50, 12, 32, 46, 28]
We should return
[1, 4, 3, 2, 0]
as P[0] = 1
because the 0
th element of A
appears at B[1]
,
and P[1] = 4
because the 1
st element of A
appears at B[4]
,
and so on.
Note:
A, B
have equal lengths in range[1, 100]
.A[i], B[i]
are integers in range[0, 10^5]
.
Companies:
Google
Related Topics:
Hash Table
Solution 1.
// OJ: https://leetcode.com/problems/find-anagram-mappings/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
unordered_map<int, int> m;
for (int i = 0; i < B.size(); ++i) m[B[i]] = i;
vector<int> ans;
for (int n : A) ans.push_back(m[n]);
return ans;
}
};
Java
class Solution {
public int[] anagramMappings(int[] A, int[] B) {
int length = A.length;
List<Integer> listB = new ArrayList<Integer>();
for (int i = 0; i < length; i++)
listB.add(B[i]);
int[] anagram = new int[length];
for (int i = 0; i < length; i++) {
int num = A[i];
int index = listB.indexOf(num);
anagram[i] = index;
listB.set(index, -1);
}
return anagram;
}
}