README ¶
762. Find Anagram Mappings
题目
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].
解题思路
见程序注释
我把问题想复杂了,以下是想复杂以后的解法。
输入
[47 34 51 47 47 34]
[47 51 34 34 47 47]
输出
[0, 2, 1, 5, 4, 3]
复杂的地方在于,输出包含了 [0,len(A)) 之间的所有整数
func anagramMappings(A, B []int) []int {
n := len(A)
indexs := make(map[int]int, n)
values := make(map[int]int, n)
res := make([]int, n)
for i := range res {
if A[i] == B[i] {
res[i] = i
continue
}
if val, ok := values[A[i]]; ok {
res[i] = val
delete(values, A[i])
} else {
indexs[A[i]] = i
}
if idx, ok := indexs[B[i]]; ok {
res[idx] = i
delete(indexs, B[i])
} else {
values[B[i]] = i
}
}
return res
}
Documentation ¶
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