LeetCode题解：Single Number I and II

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?

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?

思路：

其实是一个小技巧。一个整数和它本身异或之后得到值是0。所以初始化一个值为0的变量，让数组中的所有数与之异或，然后就可以找到这个只出现一次的数。

题目可以扩展到寻找唯一的一个只出现奇数次的数。一样的方法。

对于第二个问题，因为只能用O(1)的空间，所以技巧是对每一个位的1的个数进行计数。这样唯一的只出现一次的数用到的位将导致计数不是3的倍数。最后检查所有计数不是3倍数的位，即可恢复原来的数字。

题解：

class Solution {
public:
int singleNumber(int A[], int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.
int retval = 0;
for_each(A, A+n, [&retval](int val){retval ^= val;});
return retval;
}
};

class Solution {
public:
int singleNumber(int A[], int n) {
const size_t INTLEN = sizeof(int) * 8;
int bitCount[INTLEN];
fill(bitCount, bitCount + INTLEN, 0);

for(int i = 0; i < n; ++i)
for(size_t j = 0; j < INTLEN; ++j)
bitCount[j] += ((A[i] & (1 << j)) != 0);

int single = 0;
for(size_t j = 0; j < INTLEN; ++j)
if ((bitCount[j] % 3) == 1)
single += (1 << j);

return single;
}
};