Given two lists Aand 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 ith 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 0th element of A appears at B[1],
and P[1] = 4 because the 1st element of A appears at B[4],
and so on.
Note:
A, Bhave equal lengths in range[1, 100].A[i], B[i]are integers in range[0, 10^5].
给定两个A和B的列表，B是A的一个字母组。B是A的一个字母组，意味着B是通过随机化A中元素的顺序而制成的。
我们希望找到一个从A到B的索引映射P.映射P [i] = j意味着A中的第i个元素出现在索引为j的B中。
这些列表A和B可能包含重复项。如果有多个答案，则输出它们中的任何一个。
我们希望找到一个从A到B的索引映射P.映射P [i] = j意味着A中的第i个元素出现在索引为j的B中。
这些列表A和B可能包含重复项。如果有多个答案，则输出它们中的任何一个。
/*** @param {number[]} A* @param {number[]} B* @return {number[]}*/var anagramMappings = function (A, B) {let res = [];res.length = A.length;let m = {};for (let i = 0; i < B.length; i++) {m[B[i]] = i;}for (let i in A) {res[i] = m[A[i]];}return res;};let A = [12, 28, 46, 32, 50];let B = [50, 12, 32, 46, 28];let res = anagramMappings(A, B);console.log(res);
