You are given an array nums, where each number in the array appears either once or twice.
Return the bitwise XOR of all the numbers that appear twice in the array, or 0 if no number appears twice.
\n
Example 1:
\n\nInput: nums = [1,2,1,3]
\n\nOutput: 1
\n\nExplanation:
\n\nThe only number that appears twice in nums is 1.
Example 2:
\n\nInput: nums = [1,2,3]
\n\nOutput: 0
\n\nExplanation:
\n\nNo number appears twice in nums.
Example 3:
\n\nInput: nums = [1,2,2,1]
\n\nOutput: 3
\n\nExplanation:
\n\nNumbers 1 and 2 appeared twice. 1 XOR 2 == 3.
\n
Constraints:
\n\n1 <= nums.length <= 501 <= nums[i] <= 50nums appears either once or twice.给你一个数组 nums ，数组中的数字 要么 出现一次，要么 出现两次。
请你返回数组中所有出现两次数字的按位 XOR 值，如果没有数字出现过两次，返回 0 。
\n\n
示例 1：
\n\n输入：nums = [1,2,1,3]
\n\n输出：1
\n\n解释：
\n\nnums 中唯一出现过两次的数字是 1 。
示例 2：
\n\n输入：nums = [1,2,3]
\n\n输出：0
\n\n解释：
\n\nnums 中没有数字出现两次。
示例 3：
\n\n输入：nums = [1,2,2,1]
\n\n输出：3
\n\n解释：
\n\n数字 1 和 2 出现过两次。1 XOR 2 == 3 。
\n\n
提示：
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