760. Find Anagram Mappings (leetcode)

Find Anagram Mappings

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题目

leetcode题目

Given two lists A and 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].

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解决

1.数组A中的每个数在数组B中检查，得到index。(n为数组的大小)

• 时间复杂度为O(n * n)
• 空间复杂度为O(n)

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

2.将数组B利用unordered_map储存起来，B[i]作为key，i作为key对应的值。再遍历数组A，得到相应的index。(n为数组的大小)

• 时间复杂度为O(n)
• 空间复杂度为O(n)

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