解码异或后的排列

/* 
 * 题目思路：
 * 1. 我们通过Stream API来计算totalXor和oddXor。
 * 2. 其余步骤和Java解法一致，通过first和encoded解出perm数组。
 */
import java.util.stream.IntStream;
public class Solution {
    public int[] decode(int[] encoded) {
        int n = encoded.length + 1;
        int totalXor = IntStream.rangeClosed(1, n).reduce(0, (a, b) -> a ^ b);
        int oddXor = IntStream.range(1, n - 1).filter(i -> i % 2 == 1).map(i -> encoded[i]).reduce(0, (a, b) -> a ^ b);
        int first = totalXor ^ oddXor;
        int[] perm = new int[n];
        perm[0] = first;
        IntStream.range(0, n - 1).forEach(i -> perm[i + 1] = perm[i] ^ encoded[i]);
        return perm;
    }
}