cs106x-lecture14(Autumn 2017)-SPL实现

打卡cs106x(Autumn 2017)-lecture14

(以下皆使用SPL实现，非STL库，后续课程结束会使用STL实现)

1、min

Write a function named min that accepts a pointer to a ListNode representing the front of a linked list. Your function should return the minimum value in the linked list of integers. If the list is empty, you should throw a string exception.

Constraints: Do not construct any new ListNode objects in solving this problem (though you may create as many ListNode* pointer variables as you like). Do not use any auxiliary data structures to solve this problem (no array, vector, stack, queue, string, etc). Your function should not modify the linked list's state; the state of the list should remain constant with respect to your function.

Assume that you are using the ListNode structure as defined below:

解答：

cpp 复制代码

int min(ListNode* front) {
    int m;
    if (front == nullptr) {
        throw std::string("");
    } else {
        m = front->data;
        ListNode* current = front;
        while (current->next != nullptr) {
            current = current->next;
            if (m > current->data) {
                m = current->data;
            }
        }
    }
    return m;
}

2、countDuplicates

Write a function named countDuplicates that accepts a pointer to a ListNode representing the front of a linked list. Your function should return the number of duplicates in a sorted list. Your code should assume that the list's elements will be in sorted order, so that all duplicates will be grouped together. For example, if a variable named front points to the front of the following sequence of values, the call of countDuplicates(front) should return 7 because there are 2 duplicates of 1, 1 duplicate of 3, 1 duplicate of 15, 2 duplicates of 23 and 1 duplicate of 40:

复制代码

{1, 1, 1, 3, 3, 6, 9, 15, 15, 23, 23, 23, 40, 40}

Assume that you are using the ListNode structure as defined below:

解答：

cpp 复制代码

int countDuplicates(ListNode* front) {
    int count = 0;
    if (front == nullptr) {
        return count;
    } else {
        ListNode* current = front;
        while (current->next != nullptr) {
            if (current->data == current->next->data) {
                count++;
            }
            current = current->next;
        }
    }
    return count;
}