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Copy path1073.addNegabinary.cpp
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58 lines (52 loc) · 1.69 KB
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#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
vector<int> addNegabinary(vector<int> &arr1, vector<int> &arr2) {
return baseNeg2(getValue(arr1) + getValue(arr2));
}
int addNegabinary1(vector<int> &arr1, vector<int> &arr2) {
return getValue(arr1) + getValue(arr2);
}
private:
int getValue(const vector<int> &arr) {
int n = arr.size();
int sum = 0;
int j = 1;
for (const auto &i : arr) {
if (i == 1) {
sum += i * pow(-2, n - j);
}
j++;
}
return sum;
}
vector<int> baseNeg2(int n) {
if (n == 0)
return {0};
vector<int> ans;
while (n != 0) {
// 获取当前最低位f
if (n % -2 == 0) // 当前n为偶数,当前n的最低位为0,后续直接抹去这个最低位,除以-2即可
ans.push_back(0);
else // 当前n为奇数,当前n的最低位f为1 (tips:虽然 n % -2 可能为1或-1,但不影响,都是最低位f=1)
{
n -= 1; // 将最低位的值抹去后,使得n变为偶数,在进行除负二操作时是整除
ans.push_back(1);
}
// 将n的负二进制表达式向右移一位
n /= -2;
}
reverse(ans.begin(), ans.end()); // 除法和取余运算得到的是从低位到高位的结果,故最后要翻转
return ans;
}
};
int main() {
vector<int> arr1{1, 1, 1, 1, 1};
vector<int> arr2{1, 0, 1};
auto a = Solution().addNegabinary(arr1, arr2);
for (auto i : a) {
cout << i << " ";
}
return 0;
}