标签:arr xor XOR int Queries Li queries Subarray array
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
思路:A[i] 表示[0,i]的所有值的异或,[i, j]的异或为A[j] ^ A[i - 1], 因为x ^ x = 0.
1 class Solution { 2 public: 3 vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) { 4 for (int i = 1; i < arr.size(); ++i) { 5 arr[i] ^= arr[i - 1]; 6 } 7 vector<int> ans(queries.size()); 8 for (int i = 0; i < queries.size(); ++i) { 9 ans[i] = queries[i][0] > 0 ? arr[queries[i][1]] ^ arr[queries[i][0] - 1] : arr[queries[i][1]]; 10 } 11 return ans; 12 } 13 };
python:
1 class Solution: 2 def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]: 3 for i in range(len(arr) - 1): 4 arr[i + 1] ^= arr[i] 5 return [arr[j] ^ arr[i - 1] if i else arr[j] for i, j in queries]
标签:arr,xor,XOR,int,Queries,Li,queries,Subarray,array 来源: https://www.cnblogs.com/qinduanyinghua/p/12158469.html
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