Single Number和Single Number II

1 Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

解析:

a ^ a = 0，a ^ 0 = a

所以对所有的数取异或，其最终结果即为所求。

    public int singleNumber(int[] A) {
        int single = 0;
        for (int i : A) {
            single ^= i;
        }
        return single;
    }

2 Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

解析：

(1) a @ a @ a = 0，a @ 0 = a

只要能找到满足上述等式的操作函数@(即自定义进制为3的异或)，即可解决。

(2) a -> one，a @ a -> two，a @ a @ a -> one @ two = 0

即操作一次a，结果记录到one；操作两次，结果记录到two；操作三次时，结果记录到了one和two，则将该结果置为0.

    public int singleNumber_3(int[] A) {
        int one = 0;
        int two = 0;
        for (int carry : A) {
            while (carry != 0) {
                two ^= one & carry;
                one ^= carry;    
                carry = one & two;
            }
        }
        return one;
    }

 3 Single Number II (2)

网上流传的代码如下：

    public int singleNumber_3_2(int[] A) {
        int one = 0;
        int two = 0;
        int three = 0;
        for (int carry : A) {
            two |= one & carry;     // 这两行操作的结果是把carry加到one并进位到two。进位一般是two ^= one & carry ，这里是|，其结果是进位到two后并不继续向高位进位
            one ^= carry;           
            three = one & two;      // 在这里手动进位到three
            one &= ~three;          // one清零
            two &= ~three;          // two清零
        }
        return one;
    }

和2是等价的，由于不可能无限循环进位，2是可以写成这样的：

    public int singleNumber_3(int[] A) {
        int one = 0;
        int two = 0;
        for (int carry : A) {
            two ^= one & carry;
            one ^= carry;    
            carry = one & two;
            two ^= one & carry;
            one ^= carry; 
        }
        return one;
    }

这时候，2和3在看起来就比较像了