Hello , This is my first blog here
I’m Minezeratul
今天的每日一题为 1486. 数组异或操作
异或操作 , 即为XOR , 同为0 ,异为1 ,1^0 = 1 0^0 = 0
public static int xorOperation(int n, int start) {
if (n == 1) {
return start;
}
int res = 0 ;
for (int i = 0; i < n; i++) {
res ^= start + 2 * i;
}
return res;
}
时间复杂度为O(n)