1720:解码异或后的数组
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src/main/java/leetcode/editor/cn/DecodeXoredArray.java
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src/main/java/leetcode/editor/cn/DecodeXoredArray.java
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//未知 整数数组 arr 由 n 个非负整数组成。
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//
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// 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,a
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//rr = [1,0,2,1] 经编码后得到 encoded = [1,2,3] 。
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//
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// 给你编码后的数组 encoded 和原数组 arr 的第一个元素 first(arr[0])。
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//
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// 请解码返回原数组 arr 。可以证明答案存在并且是唯一的。
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//
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//
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//
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// 示例 1:
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//
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//
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//输入:encoded = [1,2,3], first = 1
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//输出:[1,0,2,1]
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//解释:若 arr = [1,0,2,1] ,那么 first = 1 且 encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [
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//1,2,3]
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//
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//
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// 示例 2:
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//
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//
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//输入:encoded = [6,2,7,3], first = 4
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//输出:[4,2,0,7,4]
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//
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//
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//
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//
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// 提示:
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//
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//
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// 2 <= n <= 104
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// encoded.length == n - 1
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// 0 <= encoded[i] <= 105
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// 0 <= first <= 105
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//
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// Related Topics 位运算
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// 👍 56 👎 0
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package leetcode.editor.cn;
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//1720:解码异或后的数组
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public class DecodeXoredArray {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new DecodeXoredArray().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int[] decode(int[] encoded, int first) {
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int size = encoded.length;
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int[] result = new int[size + 1];
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for (int i = 0; i < size + 1; i++) {
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if (i == 0) {
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result[i] = first;
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} else {
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result[i] = result[i - 1] ^ encoded[i - 1];
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}
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}
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return result;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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src/main/java/leetcode/editor/cn/DecodeXoredArray.md
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src/main/java/leetcode/editor/cn/DecodeXoredArray.md
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<p><strong>未知</strong> 整数数组 <code>arr</code> 由 <code>n</code> 个非负整数组成。</p>
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<p>经编码后变为长度为 <code>n - 1</code> 的另一个整数数组 <code>encoded</code> ,其中 <code>encoded[i] = arr[i] XOR arr[i + 1]</code> 。例如,<code>arr = [1,0,2,1]</code> 经编码后得到 <code>encoded = [1,2,3]</code> 。</p>
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<p>给你编码后的数组 <code>encoded</code> 和原数组 <code>arr</code> 的第一个元素 <code>first</code>(<code>arr[0]</code>)。</p>
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<p>请解码返回原数组 <code>arr</code> 。可以证明答案存在并且是唯一的。</p>
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<p> </p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>encoded = [1,2,3], first = 1
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<strong>输出:</strong>[1,0,2,1]
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<strong>解释:</strong>若 arr = [1,0,2,1] ,那么 first = 1 且 encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3]
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>encoded = [6,2,7,3], first = 4
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<strong>输出:</strong>[4,2,0,7,4]
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</pre>
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<p> </p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>2 <= n <= 10<sup>4</sup></code></li>
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<li><code>encoded.length == n - 1</code></li>
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<li><code>0 <= encoded[i] <= 10<sup>5</sup></code></li>
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<li><code>0 <= first <= 10<sup>5</sup></code></li>
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</ul>
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<div><div>Related Topics</div><div><li>位运算</li></div></div>\n<div><li>👍 56</li><li>👎 0</li></div>
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