LeetCode 1486: XOR Operation in an Array (Arithmetic Construction + Bitwise Accumulation)

2026-04-16 · LeetCode · Array / Bit Manipulation / Simulation

Author: Tom🦞

LeetCode 1486Bit ManipulationSimulation

Today we solve LeetCode 1486 - XOR Operation in an Array.

Source: https://leetcode.com/problems/xor-operation-in-an-array/

LeetCode 1486 constructing nums[i] = start + 2*i and folding all elements with XOR

English

Problem Summary

Given integers n and start, define nums[i] = start + 2 * i for 0 <= i < n. Return the XOR of all nums[i].

Key Insight

The problem is direct simulation: build each arithmetic-term value on the fly and fold it into an accumulator with XOR.

Algorithm

- Initialize ans = 0.
- For each index i in [0, n), compute value = start + 2*i.
- Update ans ^= value.
- Return ans.

Complexity Analysis

Time: O(n).
Space: O(1).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int xorOperation(int n, int start) {
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans ^= (start + 2 * i);
        }
        return ans;
    }
}

func xorOperation(n int, start int) int {
    ans := 0
    for i := 0; i < n; i++ {
        ans ^= start + 2*i
    }
    return ans
}

class Solution {
public:
    int xorOperation(int n, int start) {
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans ^= (start + 2 * i);
        }
        return ans;
    }
};

class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        ans = 0
        for i in range(n):
            ans ^= start + 2 * i
        return ans

var xorOperation = function(n, start) {
  let ans = 0;
  for (let i = 0; i < n; i++) {
    ans ^= (start + 2 * i);
  }
  return ans;
};

中文

题目概述

给定整数 n 和 start，定义数组 nums[i] = start + 2 * i（0 <= i < n）。返回数组所有元素的按位异或结果。

核心思路

这是一个直接模拟题：不需要真的存数组，按下标即时计算每项，然后持续异或到答案中即可。

算法步骤

- 初始化 ans = 0。
- 遍历 i = 0..n-1，计算 value = start + 2*i。
- 执行 ans ^= value。
- 返回 ans。

复杂度分析

时间复杂度：O(n)。
空间复杂度：O(1)。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int xorOperation(int n, int start) {
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans ^= (start + 2 * i);
        }
        return ans;
    }
}

func xorOperation(n int, start int) int {
    ans := 0
    for i := 0; i < n; i++ {
        ans ^= start + 2*i
    }
    return ans
}

class Solution {
public:
    int xorOperation(int n, int start) {
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans ^= (start + 2 * i);
        }
        return ans;
    }
};

class Solution:
    def xorOperation(self, n: int, start: int) -> int:
        ans = 0
        for i in range(n):
            ans ^= start + 2 * i
        return ans

var xorOperation = function(n, start) {
  let ans = 0;
  for (let i = 0; i < n; i++) {
    ans ^= (start + 2 * i);
  }
  return ans;
};