Leetcode 760. Find Anagram Mappings, java解法_findanagram java-优快云博客

本文链接：https://blog.youkuaiyun.com/YuanTheCoder/article/details/95327266

博客围绕给定两个列表，其中一个是另一个的变位词，要找出索引映射的问题展开。介绍了两种解法，一是暴力法，虽未处理重复元素情况但因题目允许多解可通过，时间复杂度为O(A.length ^2)；二是哈希表法，时间复杂度为O(A.length)，还提及相关Java知识点。

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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]

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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Approach #1 Brute Force, 我的解法

class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        
        int len = A.length;
        int[] ans = new int[len];

        for(int i=0; i<len; i++){
            for(int j=0;j<len;j++){
                if(A[i]==B[j]){
                    ans[i]=j;
                }
            }
        }
            
        return ans;
    }
}

分析：

虽然通过了，但是没有处理可能存在的A, B有重复元素的情况。不过题目说如有多解，允许返回任意解，可以通过。

时间复杂度： O(A.length ^2)

空间复杂度：O(A.length)

知识点：

java语言使用new 来创建数组。

dataType[] arrayRefVar = new dataType[arraySize];

或者

dataType[] arrayRefVar = {value0, value1, ..., valuek};

Approach #2 Hash Map

class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        Map<Integer, Integer> D = new HashMap();
        for (int i = 0; i < B.length; ++i)
            D.put(B[i], i);

        int[] ans = new int[A.length];
        int t = 0;
        for (int x: A)
            ans[t++] = D.get(x);
        return ans;
    }
}

分析：

时间复杂度：O(A.length)

空间复杂度： O(A.length)

知识点：

1. HashMap

Type Parameters:

K - the type of keys maintained by this map

V - the type of mapped values

Map<K,V> nameOfMap = new HashMap();

2. Method

nameOfMap.get(): Returns the value to which the specified key is mapped, or null if this map contains no mapping for the key.

nameOfMap.put(K Key, V Value) : Associates the specified value with the specified key in this map. If the map previously contained a mapping for the key, the old value is replaced.