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  • leetcode 1310. XOR Queries of a Subarray

    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]
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  • 原文地址:https://www.cnblogs.com/qinduanyinghua/p/12158469.html
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