Josephus 选举问题数组实现

Question

我正在尝试使用数组作为模拟循环链表来实现约瑟夫斯选举问题。

item 和 next 数组代表循环链表的节点。在每个第 M 个元素上，一个元素被标记为删除。下一个数组跳过标记的索引。

当数组中剩下一个元素时，我们打印出该元素和剩下的项目。

问题来自 Sedgewick C++ 算法书籍第三版。练习 3.31。

我得到的输出不正确，我没有迷路。

void array_represent_linked_list(int num_ints, int M) {
    int* item = new int[num_ints];
    int* next = new int[num_ints];

    //populated item and next array with elements
    for (int index = 0; index < num_ints; index++) {
        item[index] = index + 1;

        if (index == (num_ints - 1)) {
            next[index] = 0;
        }
        else {
            next[index] = index + 1;
        }
    }

    int nums_left = num_ints;
    int x = 0;
    int count = 0;
    int last_element_index = 0;

    while ( nums_left > 1) {

        //if current value divisible by M-2?
        if ((count % M-2) == 0) {

            if ((nums_left - 1) == 1) {

                //record the next index which is the last element
                last_element_index = next[x];

            }
            else {
                //mark for removal of element from array
                next[x] = next[next[x]];
            }

            nums_left -= 1;

        }

        //move to next element of array
        x = next[x];
        
        count++;
    }
    std::cout << item[last_element_index]<< " " <<  last_element_index<< std::endl;
    }

输出

array_represent_linked_list(9,5); //item[x] =8 , next[x] = 7

Answer 1

在编写了一个蛮力方法后，该方法查看项目数组中的每个单独元素，并在我们达到 M 的跳过因子时将每个元素设置为 0。我能够使用这段代码 sn-p 进行测试并得到正确的答案.

分解问题，每个 M-1 个元素都应该将 next[index] 设置为 next[next[index]] ，我们用 x = next[x] 遍历数组。

void array_represent_linked_list(int num_ints, int M) {
    int* item = new int[num_ints];
    int* next = new int[num_ints];

    //populated item and next array with elements
    for (int index = 0; index < num_ints; index++) {
        item[index] = index + 1;

        if (index == (num_ints - 1)) {
            next[index] = 0;
        }
        else {
            next[index] = index + 1;
        }
    }

    int nums_left = num_ints;

    //used to track elements in item array
    int index = 0;

    //used to count number of elemenets traversed in the list
    int count = 0;

    //store the return values
    int return_index = 0;
    int return_val = 0;


    while (nums_left > 1) {

        count += 1;

        if (count == (M - 1)) {

            //reset the count after reaching 4 elements
            count = 0;

            //update next element
            next[index] = next[next[index]];

            //decrease nums found
            nums_left -= 1;
        }

        //traverse to next element
        index = next[index];
    }
    return_index = index;
    return_val = item[next[index]];

    std::cout << return_index << " " << return_val << std::endl;
}