Leetcode代码学习周记——Find Anagram Mappings

原创于 2018-01-20 19:07:48 发布 · 276 阅读

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本文介绍了一种解决LeetCode上寻找字母异位词映射问题的方法。该问题要求给定两个列表A和B（B是A的字母异位词），找到从A到B的索引映射P。通过两层循环比较A和B中的元素，当找到匹配项时记录索引，最终返回索引映射数组。

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题目链接：

https://leetcode.com/problems/find-anagram-mappings/description/

题目描述：

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 Aappears 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]

P[0]
 = 1

because the 0th element of A appears at B[1], and P[1] = 4 because the 1st element ofA appears at B[4], and so on.

题解：

两层循环，若A[i] == B[j],则将设res[i] = j。

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