760. Find Anagram Mappings 查找映射

Posted 2020-10-20 Long Long Journey

 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:

1. A, B have equal lengths in range [1, 100].
2. 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。

/**
 * @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);

来自为知笔记(Wiz)