LeetCode 1310. 子数组异或查询 Java/C++ 前缀和

Java/C++ 前缀和

链接：https://leetcode-cn.com/problems/xor-queries-of-a-subarray/solution/javac-qian-zhui-he-by-oyzg-wtva/
解题思路

利用异或具有 x^x = 0；的特点
用数组encoded记录从头异或到arr[i-1]的结果
[l,r]的异或结果就等于 [1,l-1]¹的结果 因为前l-1个数都异或了2次，就抵消了

代码：
java：

class Solution {
    public int[] xorQueries(int[] arr, int[][] queries) {
        int n = arr.length;
		int m = queries.length;
		int[] encoded = new int[n+1];
		for(int i = 1; i <= n; i++) {
			encoded[i] = encoded[i-1]^arr[i-1];
		}
		int[] res = new int[m];
		for(int i = 0; i < n; i++) {
			res[i] = encoded[queries[i][0]]^encoded[queries[i][1]+1];
		}
		return res;
    }
}

class Solution {
public:
    vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) {
        int n = arr.size();
		int m = queries.size();
		vector<int> encoded(n+1);
		for(int i = 1; i <= n; i++) {
			encoded[i] = encoded[i-1]^arr[i-1];
		}
		vector<int> res(m);
		for(int i = 0; i < m; i++) {
			res[i] = encoded[queries[i][0]]^encoded[queries[i][1]+1];
		}
		return res;
    }
};

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1. 1,r ↩︎