假设玩家没有两次出现在列表中,则双
for循环非常快:
for (int i=0, i <= playerList.Count - 2, i++) for (int j=i+1, j <= playerList.Count - 1, j++) //add a new pairing of player i and j
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假设玩家没有两次出现在列表中,则双
for循环非常快:
for (int i=0, i <= playerList.Count - 2, i++) for (int j=i+1, j <= playerList.Count - 1, j++) //add a new pairing of player i and j
欢迎分享,转载请注明来源:内存溢出
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