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[Swift]LeetCode1310. 子数组异或查询 | XOR Queries of a Subarray

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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.

 

Example 1:

Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
Output: [2,7,14,8] 
Explanation: 
The binary representation of the elements in the array are:
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
The XOR values for queries are:
[0,1] = 1 xor 3 = 2 
[1,2] = 3 xor 4 = 7 
[0,3] = 1 xor 3 xor 4 xor 8 = 14 
[3,3] = 8

Example 2:

Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
Output: [8,0,4,4]

 

Constraints:

  • 1 <= arr.length <= 3 * 10^4
  • 1 <= arr[i] <= 10^9
  • 1 <= queries.length <= 3 * 10^4
  • queries[i].length == 2
  • 0 <= queries[i][0] <= queries[i][1] < arr.length

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。

 

示例 1:

输入:arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
输出:[2,7,14,8] 
解释:
数组中元素的二进制表示形式是:
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
查询的 XOR 值为:
[0,1] = 1 xor 3 = 2 
[1,2] = 3 xor 4 = 7 
[0,3] = 1 xor 3 xor 4 xor 8 = 14 
[3,3] = 8

示例 2:

输入:arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
输出:[8,0,4,4]

 

提示:

  • 1 <= arr.length <= 3 * 10^4
  • 1 <= arr[i] <= 10^9
  • 1 <= queries.length <= 3 * 10^4
  • queries[i].length == 2
  • 0 <= queries[i][0] <= queries[i][1] < arr.length

Runtime: 556 ms
Memory Usage: 25.7 MB
 1 class Solution {
 2     func xorQueries(_ arr: [Int], _ queries: [[Int]]) -> [Int] {
 3         var arr = arr
 4         var res:[Int] = [Int](repeating:0,count:queries.count)
 5         var q:[Int] = [Int]()
 6         for i in 1..<arr.count
 7         {
 8             arr[i] ^= arr[i - 1]
 9         }
10         for i in 0..<queries.count
11         {
12             q = queries[i]
13             res[i] = q[0] > 0 ? arr[q[0] - 1] ^ arr[q[1]] : arr[q[1]]
14         }
15         return res
16     }
17 }

Runtime: 540 ms
Memory Usage: 26.2 MB
1  class Solution {
2     func xorQueries(_ arr: [Int], _ queries: [[Int]]) -> [Int] {
3         var prefix = Array<Int>(repeating: 0, count: arr.count + 1)
4         for index in 0..<arr.count {
5             prefix[index + 1] = prefix[index] ^ arr[index]
6         }
7         return queries.map {prefix[$0[0]] ^ prefix[$0[1] + 1]}
8     }
9  }

 

posted @ 2020-01-05 12:12  为敢技术  阅读(353)  评论(0编辑  收藏  举报