LeetCode每日一题(1835. Find XOR Sum of All Pairs Bitwise AND)

The XOR sum of a list is the bitwise XOR of all its elements. If the list only contains one element, then its XOR sum will be equal to this element.

For example, the XOR sum of [1,2,3,4] is equal to 1 XOR 2 XOR 3 XOR 4 = 4, and the XOR sum of [3] is equal to 3.
You are given two 0-indexed arrays arr1 and arr2 that consist only of non-negative integers.

Consider the list containing the result of arr1[i] AND arr2[j] (bitwise AND) for every (i, j) pair where 0 <= i < arr1.length and 0 <= j < arr2.length.

Return the XOR sum of the aforementioned list.

Example 1:

Input: arr1 = [1,2,3], arr2 = [6,5]
Output: 0

Explanation: The list = [1 AND 6, 1 AND 5, 2 AND 6, 2 AND 5, 3 AND 6, 3 AND 5] = [0,1,2,0,2,1].
The XOR sum = 0 XOR 1 XOR 2 XOR 0 XOR 2 XOR 1 = 0.

Example 2:

Input: arr1 = [12], arr2 = [4]
Output: 4

Explanation: The list = [12 AND 4] = [4]. The XOR sum = 4.

Constraints:

• 1 <= arr1.length, arr2.length <= 105
• 0 <= arr1[i], arr2[j] <= 109

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纯数学问题
(a1 ^ a2) & (b1 ^ b2) = (a1 & b1) ^ (a1 & b2) ^ (a2 & b1) ^ (a2 & b2)
上面的等式右边是题目描述的情况， 左边是转化后的形式， 这样我们要的答案就是:
(arr1[0] ^ arr1[1] … ^ arr1[arr1.len()-1]) & (arr2[0] ^ arr2[1] … ^ arr2[arr2.len()-1])

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impl Solution {
    pub fn get_xor_sum(arr1: Vec<i32>, arr2: Vec<i32>) -> i32 {
        let x1 = arr1.into_iter().fold(0, |s, v| s ^ v);
        let x2 = arr2.into_iter().fold(0, |s, v| s ^ v);
        x1 & x2
    }
}