LeetCode Single Number and Expansion

Knowledge Point: exclusive or operation

Idea: for example, the number B ^ B = 0

A ^ 0 =

Therefore, all numbers in the array can be exclusive or, And the eventually exclusive or result is the number a that appears only once.

Code:

      ret =      i          ret ^=      ret

 

 

 

 

: One of the numbers in an array appears only once, and the other numbers appear three times. Please find the number that appears only once?

)

0 1 0 0

0 1 1 1

0 1 0 0

0 1 0 0

　

The result 0111 obtained by Sum % 3 is 7, which is the required Single Number.

      ret =      i  range(0,32         c =         d = 1 <<          j               Array[j] &                 c += 1          c % 3             ret |=      ret

 

1, The second int's I-bit is 1, which indicates that the I-bit is accumulated 2.

0 1 0 0

One 0 1 0 0

Two 0 0 0 0

==================================

0 1 1 1

One 0 0 1 1

Two 0 1 0 0

==================================

0 1 0 0

One 0 0 0 0

Two 0 0 1 1

==================================

0 1 0 0

One 0 1 0 0

Two 0 0 0 0

==================================

Code:

     one =     two =     three =      i          two |= (one &         one ^=         three = one &         two -=         one -=      one

 

 

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:Two numbers a and B appear only once in an array, and the other numbers appear twice. Can you find the numbers that appear only once?

 

      ret =      i          ret ^=     pos =      j  range(0,32          (ret>>j) & 1             pos =                  sub1,sub2 =     single1 =     single2 =            sub1 =     sub2 =      i           (Array[i]>>pos) & 1                 sub1,sub2

      ret =      i          ret ^=     pos =      j  range(0,32          (ret>>i) & 1             pos =                  sub1,sub2 =     single1 =     single2 =            sub1 =     sub2 =      i           (Array[i]>>pos) & 1                 sub1,sub2

 

 

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:One digit, a, appears only once in an array and K in other numbers. Please find the number that appears only once?

Knowledge Point: Bit operation and score Solution

① K % 2 = 0

The basic idea is similar

② K % 2 = 1

The basic idea is similar

Code:

       k % 2 ==         ret =              ret =      ret

 

 

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: Three numbers a, B, and c appear only once in an array, and the other numbers appear twice. Can you find the three numbers that appear only once?

Idea: Here Reprinted from: http://zhedahht.blog.163.com/blog/static/25411174201283084246412/

If we convert all numbers in the array to different or, the final result (as x) returns the variance or result of the numbers a, B, and c (x = a ^ B ^ c ). The other two numbers are offset in the exclusive or operation.

 

We can prove that x is not one of the three distinct numbers a, B, and c.

We use the reverse verification method to prove this. Assume that x is equal to one of a, B, and c. For example, x equals to a, that is, a = a ^ B ^ c. Therefore, B ^ c is equal to 0, that is, B is equal to c. This is in conflict with the three distinct numbers a, B, and c.

 

Since x is different from a, B, and c, x ^ a, x ^ B, and x ^ c are not equal to 0.

 

We define a function f (n). The result is to retain the last 1 in the binary representation of the number n, and change all other bits to 0. For example, decimal 6 indicates that the binary value is 0110,

Therefore, the result of f (6) is 2 (Binary 0010 ). The results of f (x ^ a), f (x ^ B), and f (x ^ c) are not equal to 0.

Next we will consider the results of f (x ^ a) ^ f (x ^ B) ^ f (x ^ c. F (x ^ a) ^ f (x ^ B) the result of ^ f (x ^ c) is certainly not 0.

So f (x ^ a) ^ f (x ^ B) ^ f (x ^ c) has at least one binary value of 1. Assume that the last digit is 1 and the second digit is m. In the result of x ^ a, x ^ B, and x ^ c, the m-bit of one or three numbers is 1.

We use the arc method to prove that it is impossible for the three numbers to have 1 in the m position.

If the m-bits of x ^ a, x ^ B, and x ^ c are both 1, the m-bit numbers of numbers a, B, and c are the opposite of the m-bit numbers of x. Therefore, the m-bit numbers of numbers a, B, and c are the same. If the m-bits of the numbers a, B, and c are both

Is 0, x = a ^ B ^ c. The m-bit of the result is 0. Since the m-th digits of x and a are both 0, the m-th digit of the result of x ^ a should be 0. Likewise, it can be proved that x ^ B and x ^ c are both 0 at the m position. This is in conflict with our assumptions.

If the m-th digits of numbers a, B, and c are both 1, x = a ^ B ^ c, the m-th digit of the result is 1. Since the m-th digits of x and a are both 1, the m-th digit of the result of x ^ a should be 0.

Likewise, it can be proved that x ^ B and x ^ c are both 0 at the m position. This is still in conflict with our assumptions.

Therefore, only one of the three numbers x ^ a, x ^ B, and x ^ c has a 1-digit m-digit. So we found the criteria that can distinguish numbers a, B, and c.

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Here we need to note that the above theoretical proof is only known in the case of a, B, c. The specific programming is somewhat different:

We first obtain x = a ^ B ^ c

We can compare x with the elements in the array again to obtain a new array, but the array feature is not changed.

For example, there are two equal numbers in the array d

After processing, the two numbers are equal, but the result is x ^ d.

 

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In this way, we can divide the array into two parts. Here we assume that the m bit of a is 1.

Then the two sub-arrays are: sub1: including a =

Sub2: Contains B, c =

 

Code:

 

      x =      i          x ^=     newArray =      j          newArray.append(x^     lastOneArrary =     ret =      i          ret ^=     pos =      j  range(0,32          (ret>>j) & 1             pos =                  sub1, sub2 =                len(sub1)%2         single1 =         single2, single3 =              single1 =         single2, single3 =              newArray =      i          pos =          j  range(0,32              (Array[i]>>j) & 1                 pos =                          newArray.append(1<<      newArray