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]
.
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;
}