leetcode-daily/2201/220108-CN.cpp

81 lines
2.3 KiB
C++

#include <vector>
#include <iostream>
#include <chrono>
/**
* 89. Gray Code
* An n-bit gray code sequence is a sequence of 2^n integers where:
* Every integer is in the inclusive range [0, 2^n - 1],
* The first integer is 0,
* An integer appears no more than once in the sequence,
* The binary representation of every pair of adjacent integers differs by exactly one bit, and
* The binary representation of the first and last integers differs by exactly one bit.
* Given an integer n, return any valid n-bit gray code sequence.
*/
class Solution {
public:
static std::vector<int> grayCode(const int n) {
std::vector<int> ret{0};
for (int i = 0; i < n; ++i) {
ret.insert(ret.end(), ret.rbegin(), ret.rend());
for (int j = 1 << i; j < (1 << (1 + i)); ++j)
ret[j] |= (1 << i);
}
return ret;
}
};
class SolutionOld {
public:
static std::vector<int> grayCode(const int n) {
int lim = 1 << n, cur = 0;
std::vector<int> ret{0};
ret.reserve(lim);
int* const vis = new int[1 + (lim >> 5)]{0x1};
for (int i = 1; i < lim; ++i) {
for (int j = 0; j < n; ++j) {
if (!(vis[(cur ^ (1 << j)) >> 5] & (1 << ((cur ^ (1 << j)) & 0x1F)))) {
ret.push_back(cur ^= (1 << j));
vis[cur >> 5] |= (1 << (cur & 0x1F));
break;
}
}
}
delete[] vis;
return ret;
}
};
class Solution0ms {
public:
static std::vector<int> grayCode(int n) {
std::vector<int> ans = {0, 1};
for (int i = 2; i <= n; i++) {
int x = (1 << (i - 1));
int idx = ans.size() - 1;
while (idx >= 0) {
int num = (ans[idx] + x);
ans.push_back(num);
idx--;
}
}
return ans;
}
};
int main() {
const int n = 30;
auto t1 = std::chrono::high_resolution_clock::now();
SolutionOld::grayCode(n);
auto t2 = std::chrono::high_resolution_clock::now();
Solution::grayCode(n);
auto t4 = std::chrono::high_resolution_clock::now();
Solution0ms::grayCode(n);
auto t5 = std::chrono::high_resolution_clock::now();
std::cout << "SolutionOld: " << std::chrono::duration_cast<std::chrono::milliseconds>(t2 - t1).count() << "ms" << std::endl;
std::cout << "Solution: " << std::chrono::duration_cast<std::chrono::milliseconds>(t4 - t2).count() << "ms" << std::endl;
std::cout << "Solution0ms: " << std::chrono::duration_cast<std::chrono::milliseconds>(t5 - t4).count() << "ms" << std::endl;
return 0;
}