leetcode 之Single Number（13）

看见这题我的第一反应是用哈希来做，不过更简洁的做法是用异或来处理，只要是偶数个都为0（0和任意数异或仍为数本身）。

  int singleNumber(int A[], int n){int x = 0;for (int i = 0; i < n; i++)x ^= A[i];return x;}

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这题的思路更加巧妙，需好好琢磨，思路如下：

int singleNumber2(int A[], int n){int one, two, three;one = two = three = 0;for (int i = 0; i < n; i++){three = two & A[i];//已经出现了两次，再出现了一次two = two | (one &A[i]);one = one | A[i];//去掉出现三次的one = one &(~three);two = two &(~three);}return one;}

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转载于:https://www.cnblogs.com/573177885qq/p/5501771.html

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