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75 lines
2.5 KiB
Markdown
75 lines
2.5 KiB
Markdown
# [1310. XOR Queries of a Subarray](https://leetcode.com/problems/xor-queries-of-a-subarray/)
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## 题目
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Given the array `arr` of positive integers and the array `queries` where `queries[i] = [Li,Ri]`, for each query `i` compute the **XOR** of elements from `Li` to `Ri` (that is, `arr[Li]xor arr[Li+1]xor ...xor arr[Ri]`). Return an array containing the result for the given `queries`.
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**Example 1:**
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```
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Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
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Output: [2,7,14,8]
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Explanation:
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The binary representation of the elements in the array are:
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1 = 0001
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3 = 0011
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4 = 0100
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8 = 1000
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The XOR values for queries are:
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[0,1] = 1 xor 3 = 2
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[1,2] = 3 xor 4 = 7
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[0,3] = 1 xor 3 xor 4 xor 8 = 14
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[3,3] = 8
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```
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**Example 2:**
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```
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Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
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Output: [8,0,4,4]
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```
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**Constraints:**
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- `1 <= arr.length <= 3 * 10^4`
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- `1 <= arr[i] <= 10^9`
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- `1 <= queries.length <= 3 * 10^4`
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- `queries[i].length == 2`
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- `0 <= queries[i][0] <= queries[i][1] < arr.length`
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## 题目大意
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有一个正整数数组 arr，现给你一个对应的查询数组 queries，其中 queries[i] = [Li, Ri]。对于每个查询 i，请你计算从 Li 到 Ri 的 XOR 值（即 arr[Li] xor arr[Li+1] xor ... xor arr[Ri]）作为本次查询的结果。并返回一个包含给定查询 queries 所有结果的数组。
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## 解题思路
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- 此题求区间异或，很容易让人联想到区间求和。区间求和利用前缀和，可以使得 query 从 O(n) 降为 O(1)。区间异或能否也用类似前缀和的思想呢？答案是肯定的。利用异或的两个性质，x ^ x = 0，x ^ 0 = x。那么有：（由于 LaTeX 中异或符号 ^ 是特殊字符，笔者用 $\oplus$ 代替异或）
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$$\begin{aligned}Query(left,right) &=arr[left] \oplus \cdots \oplus arr[right]\\&=(arr[0] \oplus \cdots \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots \oplus arr[left-1]) \oplus (arr[left] \oplus \cdots \oplus arr[right])\\ &=(arr[0] \oplus \cdots \oplus arr[left-1]) \oplus (arr[0] \oplus \cdots \oplus arr[right])\\ &=xors[left] \oplus xors[right+1]\\ \end{aligned}$$
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按照这个思路解题，便可以将 query 从 O(n) 降为 O(1)，总的时间复杂度为 O(n)。
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## 代码
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```go
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package leetcode
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func xorQueries(arr []int, queries [][]int) []int {
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xors := make([]int, len(arr))
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xors[0] = arr[0]
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for i := 1; i < len(arr); i++ {
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xors[i] = arr[i] ^ xors[i-1]
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}
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res := make([]int, len(queries))
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for i, q := range queries {
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res[i] = xors[q[1]]
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if q[0] > 0 {
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res[i] ^= xors[q[0]-1]
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}
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}
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return res
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}
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``` |