• 1310. XOR Queries of a Subarray


    package LeetCode_1310
    
    /**
     * 1310. XOR Queries of a Subarray
     * https://leetcode.com/problems/xor-queries-of-a-subarray/description/
     *
     * 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. 1 <= arr.length <= 3 * 10^4
    2. 1 <= arr[i] <= 10^9
    3. 1 <= queries.length <= 3 * 10^4
    4. queries[i].length == 2
    5. 0 <= queries[i][0] <= queries[i][1] < arr.length
     * */
    class Solution {
        /*
        * solution: prefix xor array,
        * because a^b^a = b, q[l,r] = q[0,r] ^ q[0,l-1], so q[l,r] = prefixXOR[r+1] ^ prefixXOR[l]
        * Time complexity:O(n)+O(q), Space complexity:O(n)
        * */
        fun xorQueries(arr: IntArray, queries: Array<IntArray>): IntArray? {
            val n = arr.size
            val prefixXOR = IntArray(n + 1)
            for (i in 0 until n) {
                prefixXOR[i + 1] = prefixXOR[i] xor arr[i]
            }
            val result = IntArray(queries.size)
            for (i in queries.indices) {
                val left = queries[i][0]
                val right = queries[i][1]
                result[i] = prefixXOR[right + 1] xor prefixXOR[left]
            }
            return result
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13703158.html
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