table of Contents
Second, understand the concept of
3.2.1 O (1) time complexity of determining the number is not a power of two
3.2.2 calculates a 32-bit binary representation of an integer number of 1-
3.2.3 integer A conversion is B, the number of bits required to change the bit
3.2.6 determining whether a number is an even number
I. Introduction
Bit computing in daily programming is a problem-solving tool, and this tool is often ignored by most people, today we have come together to learn about.
Note: The following code provides PHP language version only.
Second, understand the concept of
| Operators | Explanation | Examples | Remark |
| & (AND operation) | If the two corresponding bits are 1, the result of the bit is 1, otherwise 0 | 0 & 1 = 0 1 & 1 = 1 |
0 & x = 0 |
| | (OR) | Two respective bits as long as there is a 1, the bit value of 1 results | 1 | 0 = 1 0 | 0 = 0 1 | 1 = 1 |
1 | x = 1 |
| ^ (XOR operation) | If the two bits of the same value participate in operations, compared with 0, 1 otherwise | 1 ^ 1 = 0 0 ^ 0 = 0 1 ^ 0 = 1 |
If x ^ y = z, then x ^ z = y; y ^ z = x |
| - (the negation operator) | ~ Is a unary operator, is used for a binary bitwise, i.e. 0 to 1, 1 |
~0 = 1 ~1 = 0 |
|
| << (left) | For each of a number of bits of all N-bit left shift, a right complement 0 | <<1 = 3 |
|
| >> (right) | Each N-bit binary bit-right of a number, to the right end of the low are discarded, for the unsigned number, high fill 0 | 3>> = 1 |
Third, the skills and combat
3.1 Common Skills
- x & (x-1): eliminate the last bit x 1
- a ^ b ^ b = a: element deduplication
- If x ^ y = z; the ^ z = y x; z ^ y = x
3.2 combat
3.2.1 O (1) time complexity of determining the number is not a power of two
Analysis: If the number is a power of two, it must satisfy the following two conditions
1) The number is greater than 0;
2) the number of binary representation of only the highest bit is 1
Tip: Use x & (x -1) eliminated the highest level of 1, 0 if the number is a power of two, not otherwise
Code:
<?php
checkMin(3);
function checkMin($num) {
if ($num <= 0) {
return 0;
}
return intval($num & ($num - 1)) == 0 ? 1 : 0;
}3.2.2 calculates a 32-bit binary representation of an integer number of 1-
Tips: using x & (x -1) erasing a most significant bit, i.e., calculation of the number of cancellation of a number of
Code:
<?php
getINum(7);
function getINum($num) {
$count = 0;
while ($num> 0) {
$count ++;
$num = $num & ($num - 1);
}
return $count;
}3.2.3 integer A conversion is B, the number of bits required to change the bit
Analysis: number 1, i.e. the second question two integers XOR operation, problems will eventually converted to present the results of the XOR
Code:
<?php
getDiffBit(4,5);
function getDiffBit($a, $b) {
$count = 0;
if ($a == $b) {
return $count;
}
$diff = $a ^ $b;
while ($diff > 0) {
$count ++;
$diff = $diff & ($diff - 1);
}
return $count;
}3.2.4 array, a number appears only once, and the rest appear twice, once to find the emergence of elements
Analysis: a ^ b ^ b deduplication elements
<?php
$arr = array(1,2,3,1,2,3,4);
getSingle($arr);
function getSingle($arr) {
$singleItem = null;
for($i = 0; $i < count($arr); $i ++) {
$singleItem ^= $arr[$i];
}
return $singleItem;
}
3.2.5 array, two numbers appear only once, and the rest appear twice, once to find the emergence of two elements.
Analysis: The results of all elements of the array of exclusive-or result of that is, two elements appear only once. These two elements not want to wait, so the XOR result is that there must be 1 in a seat. Locate the last occurrence of 1, and according to this position, again XOR array elements, will obtain a final two elements; reuse after x ^ xor = y stars another element
Code:
<?php
$arr = array(1,1,2,2,4,5);
getSingleEle($arr);
//因为涉及到了数组的循环，时间复杂度为O(n)
function getSingleEle($arr) {
$xor = null;
$a = null;
$b = null;
//step 1:将数组元素进行异或操作,最终会得到只出现一次的两个元素的异或结果
//因为两个元素都只出现了一次，异或的结果是一定存在1元素
for ($i = 0; $i < count($arr); $i ++) {
$xor ^= $arr[$i];
}
//step 2：寻找异或结果中最右侧1所在的位置
$position = 0;
$temp = $xor;
while($temp & 1 == 0) {
$position ++;
$temp >> 1;
}
//step 3:根据step2 中1所在的位置,获取2个元素中position位置为bit为1的元素
//最终得到元素x
for ($i = 0; $i < count($arr); $i ++) {
if (($arr[$i] >> $position) & 1) {
$a ^= $arr[$i];
}
}
//step 4: 根据x ^ xor = y公式得到y
$b = $a ^ $xor;
echo $a . '_' .$b;
}3.2.6 determining whether a number is an even number
Analysis: If a number is even, its lowest level certain bits of zero.
Code:
<?php
$a = 11;
$bool = ($a & 1) == 0 ? 1 : 0;