1720. 解码异或后的数组
1720. 解码异或后的数组
题目
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 <= 10^4
encoded.length == n - 1
0 <= encoded[i] <= 10^5
0 <= first <= 10^5
题目大意
未知 整数数组 arr
由 n
个非负整数组成。
经编码后变为长度为 n - 1
的另一个整数数组 encoded
,其中 encoded[i] = arr[i] XOR arr[i + 1]
。例如,arr = [1,0,2,1]
经编码后得到 encoded = [1,2,3]
。
给你编码后的数组 encoded
和原数组 arr
的第一个元素 first
(arr[0]
)。
请解码返回原数组 arr
。可以证明答案存在并且是唯一的。
示例 1:
输入: encoded = [1,2,3], first = 1
输出:[1,0,2,1]
解释: 若 arr = [1,0,2,1] ,那么 first = 1 且 encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3]
示例 2:
输入: encoded = [6,2,7,3], first = 4
输出:[4,2,0,7,4]
提示:
2 <= n <= 10^4
encoded.length == n - 1
0 <= encoded[i] <= 10^5
0 <= first <= 10^5
解题思路
异或运算的性质:
- 自反性:
a ^ a = 0
a ^ 0 = a
- 交换性和结合性:
a ^ b = b ^ a
(a ^ b) ^ c = a ^ (b ^ c)
- 还原关系:
- 如果
c = a ^ b
,则a = c ^ b
,且b = c ^ a
。
- 如果
根据性质 3,可以解码出原始数组。
已知:encoded[i] = res[i] ^ res[i+1]
因此: res[i+1] = encoded[i] ^ res[i]
。
- 初始时,将
first
放入res
。 - 遍历
encoded
,逐步根据res[i+1] = encoded[i] ^ res[i]
解码。
复杂度分析
- 时间复杂度:
O(n)
,需要遍历一次encoded
。 - 空间复杂度:
O(n)
,结果数组res
占用线性空间。
代码
/**
* @param {number[]} encoded
* @param {number} first
* @return {number[]}
*/
var decode = function (encoded, first) {
let res = [first]; // 初始化结果数组,包含第一个元素
let prev = first; // 记录当前的前一个元素
for (let num of encoded) {
// 解码:res[i+1] = encoded[i] ^ res[i]
prev = prev ^ num;
res.push(prev);
}
return res;
};
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2433 | 找出前缀异或的原始数组 | 位运算 数组 | 🟠 | 🀄️ 🔗 |