如何使用位运算符来返回有0位

问题描述

我一直在挣扎了很长一段时间这个问题。这是这个答案的后续问题（How来划分玩家x个到2队任意多次，不同的每一次？）。

I've been struggling for quite a long time on this problem. This is a follow-up question on this answer (How to divide x number of players into 2 teams randomly multiple times, differently every time?).

所以，我的球员x个，并为每个我给 1 LT;＆LT; ñ屏蔽值。通过使用这些面具，我可以很容易地形成2个玩家在每队比赛。现在，如果玩家的总数是5，一个可能的对手可能是这样的：

So, I have x number of players, and for each I give 1 << n mask value. By using those masks I can easily form a match with 2 players in each team. Now, if total number of players is 5, one possible match could be like this:

01100 team a
00011 team b
------------
10000 player resting

或6名球员也可能是这样的：

or with 6 players it could look like this:

100010 team a
001001 team b
-------------
000100 player resting
010000 player resting

问结果
我怎样才能获得通过对比队和B队一起口罩休息的那些球员？ （我是一个总的按位小白所以code例子是很好AP preciated）

Question
How can I get those resting players by comparing team a and team b masks together? (I'm a total bitwise noob so code examples are well appreciated)

感谢

推荐答案

XOR 价值A队和B的：

var resting = a ^ b;

然后休息的球员将通过 0 标记，即：

100010 team a
001001 team b
-------------
101011 players resting

最后，遍历结果的每个位：

Finally, iterate through each bit of the result:

var totalPlayers = 6;

for (var i = 1; i <= totalPlayers; i++) {

   if ((resting & 1) === 0) {
       console.log("Player #" + i + " is resting.");
   }

   resting >>>= 1;
}

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下面就是活生生的例子：（在Java中，没有JavaScript的，但是这不是问题）

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Here's live example: http://ideone.com/Kb3XJ (in Java, not JavaScript, but that's not the issue)

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