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
A:
类似于 subarray sum. 每个地方保存之前的值。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | class Solution { public: vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) { int n = arr.size(); vector<int> V(n+1,0); for(int i = 0;i<n;i++){ V[i+1] = V[i] ^ arr[i]; } vector<int> res; for(int i =0;i<queries.size();i++){ int l = queries[i][0]; int r = queries[i][1]; res.push_back(V[r+1] ^ V[l]); } return res; } }; |
No comments:
Post a Comment