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1734. 解码异或后的排列

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。

示例 1:

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]

示例 2:

输入: encoded = [6,5,4,6]
输出: [2,4,1,5,3]

提示:

  • 3 <= n < 105
  • n 是奇数。
  • encoded.length == n - 1

题解 (Rust)

1. 题解

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
    }
}