LeetCode刷题【1310】子数组异或查询

输入：arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
输出：[2,7,14,8] 
解释：
数组中元素的二进制表示形式是：
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
查询的 XOR 值为：
[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

输入：arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
输出：[8,0,4,4]

树状数组
基本是模板套用，不过思路要转变下

原本的前缀和套用为异或操作，本质没有改变

class Solution {
    public int[] xorQueries(int[] arr, int[][] queries) {
        FenwickTree fenwickTree = new FenwickTree(arr.length);
        for (int i = 0; i < arr.length; i++) {
            fenwickTree.add(i+1, arr[i]);
        }
        int[] ans = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            ans[i] = fenwickTree.query(queries[i][0]) ^ fenwickTree.query(queries[i][1]+1);
        }
        return ans;
    }

    class FenwickTree{
        int[] arr;

        public FenwickTree(int n){
            arr = new int[n+1];
        }

        public int lowBit(int i){
            return i & ( -i );
        }

        public void add(int idx, int num){
            while (idx < arr.length){
                arr[idx] ^= num;
                idx += lowBit(idx);
            }
        }

        public int query(int idx){
            int res = 0;
            while (idx > 0 ){
                res ^= arr[idx];
                idx -= lowBit(idx);
            }
            return res;
        }
    }
}