ChalkboardXorGame [source code]
public class ChalkboardXorGame {
static
/******************************************************************************/
class Solution {
public boolean xorGame (int[] nums) {
}
}
/******************************************************************************/
public static void main(String[] args) {
ChalkboardXorGame.Solution tester = new ChalkboardXorGame.Solution();
}
}
contest没有看了, AC rate很低, 估计来不及写;
UNFINISHED
contest看到的一个很简短的答案:
class Solution(object):
def xorGame(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
x = 0
for n in nums: x ^= n
return len(nums) % 2 == 0 or x ==
有这么简单吗?
uwi:
class Solution {
public boolean xorGame(int[] nums) {
int x = 0;
for(int v : nums)x ^= v;
return x == 0 || nums.length % 2 == 0;
}
}
看起来好像真的就是这么简单;
Problem Description
We are given non-negative integers nums[i] which are written on a chalkboard. Alice and Bob take turns erasing exactly one number from the chalkboard, with Alice starting first. If erasing a number causes the bitwise XOR of all the elements of the chalkboard to become 0, then that player loses. (Also, we'll say the bitwise XOR of one element is that element itself, and the bitwise XOR of no elements is 0.)
Also, if any player starts their turn with the bitwise XOR of all the elements of the chalkboard equal to 0, then that player wins.
Return True if and only if Alice wins the game, assuming both players play optimally.
Example:
Input: nums = [1, 1, 2]
Output: false
Explanation:
Alice has two choices: erase 1 or erase 2.
If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose.
If Alice erases 2 first, now nums becomes [1, 1]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 1 = 0. Alice will lose.
Notes:
- 1 <= N <= 1000.
- 0 <= nums[i] <= 2^16.
Difficulty:Hard
Total Accepted:970
Total Submissions:2.7K
Contributor:ngoc_lam
Companies
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Related Topics
math