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

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本文介绍了一种利用位运算解决特定数学问题的方法。通过观察数组元素间的位运算特性，将复杂的问题转化为简单的位操作，实现了快速计算任意两个数组中所有元素按位与后XOR值的总和。

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

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])


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
    }
}