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]
.
找出数组A中元素在B中出现位置的索引。如果B中不存在A中元素,则返回0。
思路:这道题是map的简单应用,用map存储B中元素及对应的索引。然后遍历A中元素在map中查找即可。
class Solution { public: vector<int> anagramMappings(vector<int>& A, vector<int>& B) { unordered_map<int, int> m; vector<int> res; for (int i = 0; i < B.size(); i++) m[B[i]] = i; for (auto& a : A) { if (m.count(a)) { res.push_back(m[a]); } else { res.push_back(0); } } return res; } }; // 6 ms