给你一个整数数组 perm
,它是前 n
个正整数的排列,且 n
是个 奇数 。
它被加密成另一个长度为 n - 1
的整数数组 encoded
,满足 encoded[i] = perm[i] XOR perm[i + 1]
。比方说,如果 perm = [1,3,2]
,那么 encoded = [2,1]
。
给你 encoded
数组,请你返回原始数组 perm
。题目保证答案存在且唯一。
输入: encoded = [3,1] 输出: [1,2,3] 解释: 如果 perm = [1,2,3] ,那么 encoded = [1 XOR 2,2 XOR 3] = [3,1]
输入: encoded = [6,5,4,6] 输出: [2,4,1,5,3]
3 <= n < 105
n
是奇数。encoded.length == n - 1
impl Solution {
pub fn decode(encoded: Vec<i32>) -> Vec<i32> {
let n = encoded.len() + 1;
let m = n / 2;
let mut perm = vec![0; n];
perm[m] = m as i32 + 1;
for i in 0..m {
perm[m] ^= ((i + 1) ^ (n - i)) as i32;
if i % 2 == 0 {
perm[m] ^= encoded[i] ^ encoded[n - 2 - i];
}
}
for i in 1..=m {
perm[m - i] = encoded[m - i] ^ perm[m - i + 1];
perm[m + i] = encoded[m + i - 1] ^ perm[m + i - 1];
}
perm
}
}