先决条件：汉明码
给定一个消息位，其形式为msgBit []数组，则任务是查找给定消息位的汉明码。
例子：
Input: S = “0101”
Output:
Generated codeword:
r1 r2 m1 r4 m2 m3 m4
0 1 0 0 1 0 1
Explanation:
Initially r1, r2, r4 is set to ‘0’.
r1 = Bitwise XOR of all bits position that has ‘1’ in its 0th-bit position.
r2 = Bitwise XOR of all bits that has ‘1’ in its 1st-bit position.
r3 = Bitwise XOR of all bits that has ‘1’ in its 2nd-bit position.
Input: S = “0111”
Output:
Generated codeword:
r1 r2 m1 r4 m2 m3 m4
0 0 0 1 1 1 1
的方法：思想是首先找到其可通过同时2 r为小于每次初始化被1个R，然后通过递增1被发现冗余位的数目(M + R + 1)，其中m是比特数在输入消息中。请按照以下步骤解决问题：
- 将r初始化为1，然后将其递增1，直到2 r小于m + r + 1 。
- 初始化大小为r + m的向量hammingCode ，该向量将为输出消息的长度。
- 1和设置汉明码[2 I –通过遍历从i = 0至r初始化-1冗余比特的所有的位置– 1] = -1 。然后按0 <= j <(r + m)的顺序将输入消息位放在hammingCode [j]不为-1的所有位置。
- 用0初始化一个变量one_count来存储1的数量，然后从i = 0遍历到(r + m – 1) 。
- 如果当前位即汉明码[i]不-1然后找到的消息位含有组位在日志2 =由1如果递增one_count第(i + 1)通过从j个横动位置i + 2到R + M (j＆(1 << x))不为0，而hammingCode [j – 1]为1 。
- 如果对于索引i ， one_count是偶数，则设置hammingCode [i] = 0，否则设置hammingCode [i] = 1 。
- 遍历后，将hammingCode []向量打印为输出消息。
下面是上述方法的实现：
C
// C program for the above approach
#include
#include
// Store input bits
int input[32];
// Store hamming code
int code[32];
int ham_calc(int, int);
void solve(int input[], int);
// Function to calculate bit for
// ith position
int ham_calc(int position, int c_l)
{
int count = 0, i, j;
i = position - 1;
// Traverse to store Hamming Code
while (i < c_l) {
for (j = i; j < i + position; j++) {
// If current boit is 1
if (code[j] == 1)
count++;
}
// Update i
i = i + 2 * position;
}
if (count % 2 == 0)
return 0;
else
return 1;
}
// Function to calculate hamming code
void solve(int input[], int n)
{
int i, p_n = 0, c_l, j, k;
i = 0;
// Find msg bits having set bit
// at x'th position of number
while (n > (int)pow(2, i) - (i + 1)) {
p_n++;
i++;
}
c_l = p_n + n;
j = k = 0;
// Traverse the msgBits
for (i = 0; i < c_l; i++) {
// Update the code
if (i == ((int)pow(2, k) - 1)) {
code[i] = 0;
k++;
}
// Update the code[i] to the
// input character at index j
else {
code[i] = input[j];
j++;
}
}
// Traverse and update the
// hamming code
for (i = 0; i < p_n; i++) {
// Find current position
int position = (int)pow(2, i);
// Find value at current position
int value = ham_calc(position, c_l);
// Update the code
code[position - 1] = value;
}
// Print the Hamming Code
printf("\nThe generated Code Word is: ");
for (i = 0; i < c_l; i++) {
printf("%d", code[i]);
}
}
// Driver Code
void main()
{
// Given input message Bit
input[0] = 0;
input[1] = 1;
input[2] = 1;
input[3] = 1;
int N = 4;
// Function Call
solve(input, N);
} C++
// C++ program for the above approach
#include
using namespace std;
// Function to generate hamming code
vector generateHammingCode(
vector msgBits, int m, int r)
{
// Stores the Hamming Code
vector hammingCode(r + m);
// Find positions of redundant bits
for (int i = 0; i < r; ++i) {
// Placing -1 at redundant bits
// place to identify it later
hammingCode[pow(2, i) - 1] = -1;
}
int j = 0;
// Iterate to update the code
for (int i = 0; i < (r + m); i++) {
// Placing msgBits where -1 is
// absent i.e., except redundant
// bits all postions are msgBits
if (hammingCode[i] != -1) {
hammingCode[i] = msgBits[j];
j++;
}
}
for (int i = 0; i < (r + m); i++) {
// If current bit is not redundant
// bit then continue
if (hammingCode[i] != -1)
continue;
int x = log2(i + 1);
int one_count = 0;
// Find msg bits containing
// set bit at x'th position
for (int j = i + 2;
j <= (r + m); ++j) {
if (j & (1 << x)) {
if (hammingCode[j - 1] == 1) {
one_count++;
}
}
}
// Generating hamming code for
// even parity
if (one_count % 2 == 0) {
hammingCode[i] = 0;
}
else {
hammingCode[i] = 1;
}
}
// Return the generated code
return hammingCode;
}
// Function to find the hamming code
// of the given message bit msgBit[]
void findHammingCode(vector& msgBit)
{
// Message bit size
int m = msgBit.size();
// r is the number of redundant bits
int r = 1;
// Find no. of redundant bits
while (pow(2, r) < (m + r + 1)) {
r++;
}
// Generating Code
vector ans
= generateHammingCode(msgBit, m, r);
// Print the code
cout << "Message bits are: ";
for (int i = 0; i < msgBit.size(); i++)
cout << msgBit[i] << " ";
cout << "\nHamming code is: ";
for (int i = 0; i < ans.size(); i++)
cout << ans[i] << " ";
}
// Driver Code
int main()
{
// Given message bits
vector msgBit = { 0, 1, 0, 1 };
// Function Call
findHammingCode(msgBit);
return 0;
} 输出：
The generated Code Word is: 0001111
时间复杂度： O((M + R) 2 )，其中M是输入消息中的位数，R是冗余位数
辅助空间： O(M + R)
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