leetcode-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?

一个trivial的解法

class Solution {
public:
int singleNumber(int A[], int n) {
map<int,int> arr;
for(int i=0;i<n;++i)
{
map<int,int>::iterator it;
it=arr.find(A[i]);
if(it==arr.end())arr.insert(pair<int,int>(A[i],1));
else (it->second)++;
}
for(map<int,int>::iterator it=arr.begin();it!=arr.end();++it)
{
if((it->second)!=3)return it->first;
}

}
};

class Solution {
public:
int singleNumber(int A[], int n) {
int one=0;
int two=0;
for(int i=0;i<n;i++)
{
one=(one^A[i])&~two;
two=(two^A[i])&~one;
}
return one;

}
};

前面有一道题single number I 中也是用的位操作，那道题是出现2次的元素，这道是3次，所以简单的用一个变量异或显然不行了

转载一个说明：

The code seems tricky and hard to understand at first glance. However, if you consider the problem in Boolean algebra form, everything becomes clear.

What we need to do is to store the number of '1's of every bit. Since each of the 32 bits follow the same rules, we just need to consider 1 bit. We know a number appears 3 times at most, so we need 2 bits to store that. Now we have 4 state, 00, 01, 10 and 11,
but we only need 3 of them.

In this solution, 00, 01 and 10 are chosen. Let 'ones' represents the first bit, 'twos' represents the second bit. Then we need to set rules for 'ones' and 'twos' so that they act as we hopes. The complete loop is 00->10->01->00(0->1->2->3/0).

For 'ones', we can get 'ones = ones ^ A[i]; if (twos == 1) then ones = 0', that can be tansformed to 'ones = (ones ^ A[i]) & ~twos'.

Similarly, for 'twos', we can get 'twos = twos ^ A[i]; if (ones* == 1) then twos = 0' and 'twos = (twos ^ A[i]) & ~ones'. Notice that 'ones*' is the value of 'ones' after calculation, that is why twos is calculated
later.

Here is another example. If a number appears 5 times at most, we can write a program using the same method. Now we need 3 bits and the loop is 000->100->010->110->001. The code looks like this:

int singleNumber(int A[], int n) {
int na = 0, nb = 0, nc = 0;
for(int i = 0; i < n; i++){
nb = nb ^ (A[i] & na);
na = (na ^ A[i]) & ~nc;
nc = nc ^ (A[i] & ~na & ~nb);
}
return na & ~nb & ~nc;
}

Or even like this:

int singleNumber(int A[], int n) {
int twos = 0xffffffff, threes = 0xffffffff, ones = 0;
for(int i = 0; i < n; i++){
threes = threes ^ (A[i] & twos);
twos = (twos ^ A[i]) & ~ones;
ones = ones ^ (A[i] & ~twos & ~threes);
}
return ones;
}

这是原链接

Statement of our problem: "Given an array of integers, every element appears k (k >1) times except for one, which appears p times(p>=1, p % k != 0). Find that single one."

As others pointed out, in order to apply the bitwise operations, we should rethink how integers are represented in computers -- by bits. To start, let's consider only one bit for now. Suppose we have an array of 1-bit numbers (which can only be 0 or 1), we'd
like to count the number of '1's in the array such that whenever the counted number of '1' reaches a certain value, say k, the count returns to zero and starts over (In case you are curious, this k will be the same as the one in the problem statement above).
To keep track of how many '1's we have encountered so far, we need a counter. Suppose the counter has m bits in binary form: xm, ..., x1 (from most significant bit to least significant bit). We can conclude at least the following four properties of the counter:

There is an initial state of the counter, which for simplicity is zero;
For each input from the array, if we hit a '0', the counter should remain unchanged;
For each input from the array, if we hit a '1', the counter should increase by one;
In order to cover k counts, we require 2^m >= k, which implies m >= logk.

Here is the key part: how each bit in the counter (x1 to xm) changes as we are scanning the array. Note we are prompted to use bitwise operations. In order to satisfy the second property, recall what bitwise operations will not change the operand if the other
operand is 0? Yes, you got it: x = x | 0 and x = x ^ 0.

Okay, we have an expression now: x = x | i or x = x ^ i, where i is the scanned element from the array. Which one is better? We don't know yet. So, let's just do the actual counting:

At the beginning, all bits of the counter is initialized to zero, i.e., xm = 0, ..., x1 = 0. Since we are gonna choose bitwise operations that guarantees all bits of the counter remain unchanged if we hit '0's, the counter will be 0 until we hit the first '1'
in the array. After we hit the first '1', we got: xm = 0, ...,x2 = 0, x1 = 1. Let's continue until we hit the second '1', after which we have: xm = 0, ..., x2 = 1, x1 = 0. Note that x1 changes from 1 to 0. For x1 = x1 | i, after the second count, x1 will still
be 1. So it's clear we should use x1 = x1 ^ i. What about x2, ..., xm? The idea is to find the condition under which x2, ..., xm will change their values. Take x2 as an example. If we hit a '1', and we need to change the value of x2, what must the value of
x1 be before we do the change? The answer is: x1 must be 1 otherwise we shouldn't change x2 because changing x1 from 0 to 1 will do the job. So x2 will change only if x1 and i are both 1, or mathematically, x2 = x2 ^ (x1 & i). Similarly xm will change only
when xm-1, ..., x1 and i are all 1: xm = xm ^ (xm-1 & ... & x1 & i); Bingo, we've found the bitwise operations!

However, you may notice that the bitwise operations found above will count from 0 until 2^m - 1, instead of k. If k < 2^m - 1, we need some "cutting" mechanism to reinitialize the counter to 0 when the count reaches k. To this end, we apply bitwise AND to xm,...,
x1 with some variable called mask, i.e., xm = xm & mask, ..., x1 = x1 & mask. If we can make sure the mask will be 0 only when the count reaches k and be 1 for all other count cases, then we are done. How do we achieve that? Try to think what distinguishes
the case with k count from all other count cases. Yes, it's the count of '1's! For each count, we have unique values for each bit of the counter, which can be regarded as its state. If we write k in its binary form: km,..., k1. we can construct the mask as
follows:

mask = ~(x1' & x2' & ... xm'), where xj' = xj if kj = 1 and xj' = ~xj if kj = 0 (j = 1 to m).

Let's do some examples:

k = 3: k1 = 1, k2 = 1, mask = ~(x1 & x2);

k = 5: k1 = 1, k2 = 0, k3 = 1, mask = ~(x1 & ~x2 & x3);

In summary, our algorithm will go like this:

for (int i : array) {

xm ^= (xm-1 & ... & x1 & i);

xm-1 ^= (xm-2 & ... & x1 & i);

.....

x1 ^= i;

mask = ~(x1' & x2' & ... xm') where xj' = xj if kj = 1 and xj' = ~xj if kj = 0 (j = 1 to m).

xm &= mask;

......

x1 &= mask;

}

Now it's time to generalize our results from 1-bit number case to 32-bit integers. One straightforward way would be creating 32 counters for each bit in the integer. You've probably already seen this in other posted codes. But if we take advantage of bitwise
operations, we may be able to manage all the 32 counters "collectively". By saying "collectively" we mean using m 32-bit integers instead of 32 m-bit counters, where m is the minimum integer that satisfies m >= logk. The reason is that bitwise operations apply
only to each bit so operations on different bits are independent of each other(kind obvious, right?). This allows us to group the corresponding bits of the 32 counters into one 32-bit integer. Since each counter has m bits, we end up with m 32-bit integers.
Therefore, in the algorithm developed above, we just need to regard x1 to xm as 32-bit integers instead of 1-bit numbers and we are done. Easy, hum?

The last thing is what value we should return, or equivalently which one of x1 to xm will equal the single element. To get the correct answer, we need to understand what the m 32-bit integers x1 to xm represent. Take x1 as an example. x1 has 32 bits and let's
label them as r (r = 1 to 32), After we are done scanning the input array, the value for the r-th bit of x1 will be determined by the r-th bit of all the elements in the array (more specifically, suppose the total count of '1' for the r-th bit of all the elements
in the array is q, q' = q % k and in its binary form: q'm,...,q'1, then by definition the r-th bit of x1 will equal q'1). Now you can ask yourself this question: what does it imply if the r-th bit of x1 is '1'?

The answer is to find what can contribute to this '1'. Will an element that appears k times contribute? No. Why? Because for an element to contribute, it has to satisfy at least two conditions at the same time: the r-th bit of this element is '1' and the number
of appearance of this '1' is not an integer multiple of k. The first condition is trivial. The second comes from the fact that whenever the number of '1' hit is k, the counter will go back to zero, which means the corresponding bit in x1 will be set to 0.
For an element that appears k times, it's impossible to meet these two conditions simultaneously so it won't contribute. At last, only the single element which appears p (p%k != 0) times will contribute. If p > k, then the first k*[p/k] (denotes the integer
part of p/k) single elements won't contribute either. Then we can always set p' = p % k and say the single element appears effectively p' times.

Let's write p' in its binary form: p'm, ..., p'1. (note that p' < k, so it will fit into m bits). Here I claim the condition for x1 to equal the single element is p'1 = 1. Quick proof: if the r-th bit of x1 is '1', we can safely say the r-th bit of the single
element is also '1'. We are left to prove that if the r-th bit of x1 is '0', then the r-th bit of the single element can only be '0'. Just suppose in this case the r-th bit of the single element is '1', let's see what will happen. At the end of the scan, this
'1' will be counted p' times. If we write p' in its binary form: p'm, ..., p'1, then by definition the r-th bit of x1 will equal p'1, which is '1'. This contradicts with the presumption that the r-th bit of x1 is '0'. Since this is true for all bits in x1,
we can conclude x1 will equal the single element if p'1 = 1. Similarly we can show xj will equal the single element if p'j = 1(j = 1 to m). Now it's clear what we should return. Just express p' = p % k in its binary form, and return any of the corresponding
xj as long as p'j = 1.

In total, the algorithm will run in O(n*logk) time and O(logk) space.