1310. XOR Queries of a Subarray ## 思路 先建PrefixSum的表, 再query差值 ex. arr = [a1, a2, a3, ...] prefix = [0, a1, a1^a2, a1^a2^a3, ...] a2^a3 = (a1^a2^a3) ^ a1 ## Code ```python class Solution: def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]: prefix = [0] for num in arr: prefix.append(prefix[-1] ^ num) res = [] for left, right in queries: res.append(prefix[right+1] ^ prefix[left]) return res ``` -- https://i.imgur.com/kyBhy6o.jpeg -- ※ 發信站: 批踢踢實業坊(web-ptt.tw), 來自: 185.213.82.238 (臺灣) ※ 文章網址: https://web-ptt.tw/Marginalman/M.1726202917.A.F41