760. Find Anagram Mappings

 

Given two lists and B, and B is an anagram of AB 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 Aappears 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].

 

这道题给了我们两个数组A和B,说是A和B中的数字都相同,但是顺序不同,有点类似错位词的感觉。让我们找出数组A中的每个数字在数组B中的位置。这道题没有太大的难度,用个HashMap建立数组B中的每个数字和其位置之间的映射,然后遍历数组A,在HashMap中查找每个数字的位置即可,参见代码如下:

 

class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        vector<int> res;
        unordered_map<int, int> m;
        for (int i = 0; i < B.size(); ++i) m[B[i]] = i;
        for (int num : A) res.push_back(m[num]);
        return res;
    }
};

 

类似题目:

Find All Anagrams in a String

Valid Anagram

Anagrams

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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