Question

我检查了SO，看到了两个处理相同问题陈述的问题。但是，它们都没有包含我正在寻找的东西。如果具有 n 元素的数组包含一个超过 n / 2 次的元素，使用分而治之策略。

我基于以下事实开发了我认为是一种有效的解决方案：基本情况是一个子数组，其中一个元素（显然）是子数组中的主要元素。然后我继续比较子阵列中的这些多数元素，询问它们最终是否超过 n / 2 次。

有关详细信息，请参阅https://classes.soe.ucsc.edu/cmps102/Fall01/solutions4.pdf（问题4）

我为这个问题编写了两个不同的解决方案，一个使用非常天真的O（n ^ 2）算法来检查我的解决方案是否适用于所有情况，还有一个尝试实现上面链接中描述的算法。

在主要内部，我强调测试我的解决方案与明显正确但天真的解决方案。但是，运行此操作后，我收到SIGABRT（信号6）错误。我的调试器告诉我检查malloc，对象可能在被释放后被修改。

现在我无法判断我的解决方案是否正确。我真的不知道发生了什么错误的分配，我对C ++来说相对较新。

代码如下：

#include <algorithm> #include <iostream> #include <vector> using std::vector; int get_majority_element(vector<int> &a, int left, int right) { int m; int majority_left, majority_right; // majority element in either sub array int count_right = 0, count_left = 0; // count for above int leftt, rightt; // endpoints if (a.size() == 1) { return a[0]; } else { m = (left + right)/2; // calculate mid point vector<int> b_left(m); vector<int> b_right(right - m); // get left sub array for (int i = 0; i < m; i++) { b_left[i] = a[i]; } // set new endpoints leftt = 0; rightt = m; majority_left = get_majority_element(b_left, leftt, rightt); for (int i = 0; i < right - m + 1; i++) { b_right[i] = a[m+i]; } leftt = m; rightt = right - m + 1; majority_right = get_majority_element(b_right, leftt, rightt); // if there is a majority element, count its frequency if (majority_left != -1) { for (int i = 0; i < a.size(); i++) { if (a[i] == majority_left) count_left++; } } if (majority_right != -1) { for (int i = 0; i < a.size(); i++) { if (a[i] == majority_right) count_right++; } } // if both elements in sub arrays are majority and they're different, there is no majority element if (count_left == count_right && majority_left != majority_right) { return -1; } // check if either sub array has a majority element that occurs more than n/2 times else if (count_right > count_left && count_right > a.size()/2) { return majority_right; } else if (count_left > count_right && count_left > a.size()/2){ return majority_left; } } return -1; } int get_majority_fast(vector<int> &a, int left, int right){ std::reverse(a.begin(),a.end()); int current = 0; int count; for (int i = 0; i < a.size(); i++) { current = a[i]; count = 0; for (int j = 0; j < a.size(); j++) { if (a[j] == current) count ++; } if (count > a.size()/2) return 1; } return -1; } int main() { // std::cin >> n; // vector<int> a(n); // for (size_t i = 0; i < a.size(); ++i) { // std::cin >> a[i]; // } // std::cout << (get_majority_fast(a, 0, a.size()) != -1) << '\n'; while(true){ int one, two; int n = rand() % 100 ; std::cout << n << "\n"; vector<int> a; for (int i = 0; i < n; ++i) { a.push_back(rand() % 100); } one = get_majority_element(a, 0, a.size()); two = get_majority_fast(a, 0, a.size() != -1); if (one != two) { std::cout << "Wrong answer: " << one << ' ' << two << "\n"; break; } else { std::cout << "OK\n"; } } }

Answer 1

其中一个循环中实际上存在越界错误。修正循环如下：

for (int i = m; i < right - m ; i++) {
    b_right.at(m-i) = a.at(m + i);
}

我认为奇怪的SIGABRT错误是由于一些神秘的东西。猜猜我学会了使用x.at()

另外，我还有其他一些错误。完整更正的代码如下：

#include <algorithm>
#include <iostream>
#include <vector>


using std::vector;

int get_majority_element(vector<int> &a, int left, int right) {


    int m;
    int majority_left = -1, majority_right = -1;      // majority element in either sub array
    int count_right = 0, count_left = 0;    // count for above
    int new_left, new_right;                      // endpoints

    if (a.size() == 1) {
        return a[0];
    }
    else {


        m = (a.size())/2;  // calculate mid point

        vector<int> b_left(m);
        vector<int> b_right(right - m);


        // get left sub array
        for (int i = 0; i < m; i++) {
            b_left.at(i) = a.at(i);
        }

        for (int i = 0; i < a.size() - m ; i++) {
            b_right.at(i) = a.at(i + m);
        }

        // set new endpoints
        new_left = 0;
        new_right = m;
        majority_left = get_majority_element(b_left, new_left, new_right);


        new_left = m;
        new_right = right - m;

        majority_right = get_majority_element(b_right, new_left, new_right);


        // if there is a majority element, count its frequency

        if (majority_left != -1) {
            for (int i = 0; i < a.size(); i++) {
                if (a[i] == majority_left)
                    count_left++;
            }
        }

        if (majority_right != -1) {
            for (int i = 0; i < a.size(); i++) {
                if (a[i] == majority_right)
                    count_right++;
            }
        }


        // if both elements in sub arrays are majority and they're different, there is no majority element
        if (count_left == count_right && majority_left != majority_right) {
            return 0;
        }
        else if (count_left == count_right)
            return majority_left;
        // check if either sub array has a majority element that occurs more than n/2 times
        else if (count_right > count_left && count_right > a.size()/2) {
                return majority_right;
        }
        else if (count_left > count_right && count_left > a.size()/2){
            return majority_left;
        }
            // majority element in sub arrays isn't majority in array
        else
            return 0;
    }

}


int get_majority_fast(vector<int> &a, int left, int right){

    std::reverse(a.begin(),a.end());

    int current = 0;
    int count;
    for (int i = 0; i < a.size(); i++) {
        current = a[i];
        count = 0;
        for (int j = 0; j < a.size(); j++) {
            if (a[j] == current)
                count ++;
        }
        if (count > a.size()/2)
            return 1;
    }
    return -1;
}

int main() {

    int n;
    std::cin >> n;
    vector<int> a(n);

    for (size_t i = 0; i < a.size(); ++i) {
        std::cin >> a[i];
    }

    int result;
    int out;

    result = get_majority_element(a, 0, a.size());

    if (result != 0)
        out = 1;
    else
        out = 0;
    std::cout << out << '\n';


}

我确信我可以让它变得更漂亮但是现在我不想听到这个问题。

SIGABRT（信号6）使用除法和征服

1 个答案: