LeetCode 760. Find Anagram Mappings

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, B have equal lengths in range [1, 100].
A[i], B[i] are integers in range [0, 10^5].

题目大意：给两个数组A和B，B是A的同字母异序词，返回一个等长数组P，其中P[i] = j，表示A的第i个元素在B的第j个元素处，如果有多个答案，返回一个即可～

分析：设置一个二维数组v，将数字i对应的在B数组中的下标放在v[i]中，这样遍历A数组，每次取出一个v[A[i]]放入ans数组对应的i下标处即可～

class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        int len = A.size();
        vector<int> ans(len), hash(len), v[100001];
        for (int i = 0; i < len; i++) v[B[i]].push_back(i);
        for (int i = 0; i < len; i++) {
            ans[i] = v[A[i]].back();
            v[A[i]].pop_back();
        }
        return ans;
    }
};

class Solution {

public:

vector<int> anagramMappings(vector<int>& A, vector<int>& B) {

int len = A.size();

vector<int> ans(len), hash(len), v[100001];

for (int i = 0; i < len; i++) v[B[i]].push_back(i);

for (int i = 0; i < len; i++) {

ans[i] = v[A[i]].back();

v[A[i]].pop_back();

}

return ans;

}

};

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