LeetCode 760. Find Anagram Mappings
原题链接在这里:https://leetcode.com/problems/find-anagram-mappings/description/
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 i
th 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 0
th element of A
appears at B[1]
, and P[1] = 4
because the 1
st 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]
.
题解:
用map保存B的元素与对应位置. 再用A的元素在map中找对应位置.
Time Complexity: O(A.length).
Space: O(1), regardless res.
AC Java:
class Solution {
public int[] anagramMappings(int[] A, int[] B) {
if(A == null || B == null || A.length != B.length){
throw new IllegalArgumentException("Invalid input.");
} HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
for(int i = 0; i<B.length; i++){
hm.put(B[i], i);
} int [] res = new int[A.length];
for(int i = 0; i<A.length; i++){
res[i] = hm.get(A[i]);
} return res;
}
}
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