LWC 66: 760. Find Anagram Mappings
传送门：760. Find Anagram Mappings
Problem:
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 B1, and P1 = 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].
思路：
很简单，用Map记录B中的值和index，然后映射到A。
代码如下：
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;
}当然你也可以暴力搜索一波，代码如下：
public int[] anagramMappings(int[] A, int[] B) {
int n = A.length;
int[] ans = new int[n];
boolean[] vis = new boolean[n];
for (int i = 0; i < n; ++i) {
int find = -1;
for (int j = 0; j < n; ++j) {
if (!vis[j]) {
if (A[i] == B[j]) {
vis[j] = true;
find = j;
break;
}
}
}
ans[i] = find;
}
return ans;
}Python版本：
def anagramMappings(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: List[int]
"""
n = len(A)
map = {}
for i, v in enumerate(B):
map[v] = i
ans = [0] * n
for i, v in enumerate(A):
ans[i] = map[v]
return ans
