QUESTION:
There is a hidden integer array arr that consists of n non-negative integers.
It was encoded into another integer array encoded of length n - 1, such that encoded[i] = arr[i] XOR arr[i + 1]. For example, if arr = [1,0,2,1], then encoded = [1,2,3].
You are given the encoded array. You are also given an integer first, that is the first element of arr, i.e. arr[0].
Return the original array arr. It can be proved that the answer exists and is unique.
Example 1:
Input: encoded = [1,2,3], first = 1
Output: [1,0,2,1]
Explanation: If arr = [1,0,2,1], then first = 1 and encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3]
Example 2:
Input: encoded = [6,2,7,3], first = 4
Output: [4,2,0,7,4]
Constraints:
2 <= n <= 104
encoded.length == n - 1
0 <= encoded[i] <= 105
0 <= first <= 105
EXPLANATION:
这道题目其实是一个算术题. 只要知道一个规律就可以了:
x xor a = b
b xor a = x
既一个数,连续异或两次同一个数,结果还是自己. 通过这个规律,我们就可以反向推出最终的结果.
SOLUTION:
class Solution {
fun decode(encoded: IntArray, first: Int): IntArray {
var result : IntArray = IntArray(encoded.size+1)
if (result.size>0) result[0] = first
for (i in encoded.indices) result[i+1] = encoded[i] xor result[i]
return result
}
}