leetcode 760. Find Anagram Mappings

最新推荐文章于 2025-08-15 12:05:50 发布

mary441

最新推荐文章于 2025-08-15 12:05:50 发布

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本文介绍了一种在两个互为字母异位词的数组间建立索引映射的方法。通过示例说明如何找到第一个数组在第二个数组中的对应元素位置，并提供两种实现方案：一种是直接遍历查找，另一种是利用哈希映射提高效率。

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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]

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].

在另一数组中找到前一数组的元素的索引，并将索引生成返回新的数组

class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        int length = A.length;
        int[] P = new int[length];
        for(int i=0;i<length;i++){
            for(int j=0;j<length;j++){
                if(A[i] == B[j]){
                    P[i]=j;
                    break;
                }
            }
        }
        return P;
    }
}

优秀解法：利用map一一对应

  public int[] anagramMappings(int[] A, int[] B) {
        int n = A.length;
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            map.put(B[i], i);
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = map.get(A[i]);
        }
        return ans;
    }