760. Find Anagram Mappings 尋找映射

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 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].

給定兩個A和B的列表，B是A的一個字母組。B是A的一個字母組，意味著B是通過隨機化A中元素的順序而製成的。
我們希望找到一個從A到B的索引映射P.映射P [i] = j意味著A中的第i個元素出現在索引為j的B中。
這些列表A和B可能包含重複項。如果有多個答案，則輸出它們中的任何一個。

 

  
/**
  
 * @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);
 

來自為知筆記(Wiz)

760. Find Anagram Mappings 尋找映射