英文:
Finding an Element in a LinkedList - C++
问题
我正在尝试编写一个函数 find(Node* node, int valueInput)，该函数从给定的节点开始，返回具有值 valueInput 的节点的索引，如果 valueInput 不存在则返回 -1。
以下是代码：
#include <iostream>class Node {public:int value;Node* next = NULL;};int find(Node* node, int valueInput){Node* tempNode = node;int count = 0;while (tempNode != NULL){if (tempNode->value == valueInput){return count;}++count;tempNode = tempNode->next;}return -1;}int main(){int valueInput, currentValue;char arrow;Node* node1, * previous, * head;std::cin >> valueInput;std::cin >> currentValue;node1 = new Node();node1->value = currentValue;node1->next = NULL;previous = node1;head = node1;while (std::cin){std::cin >> arrow;std::cin >> currentValue;node1 = new Node();node1->value = currentValue;previous->next = node1;previous = node1;}std::cout << find(head, valueInput);}
目前，我的程序始终返回 -1。
示例输入和输出：
示例输入：
5
1->2->5
示例输出：
2
英文:
I am trying to write a function find(Node* node, int valueInput) that starts at the given node and returns the index of the node with the value valueInput, and return -1 if valueInput does not exist.
Here is the code
#include <iostream>class Node {public:int value;Node* next = NULL;};int find(Node* node, int valueInput){Node* tempNode = node;int count = 0;while (tempNode != NULL){if (tempNode->value == 0){break;}++count;tempNode = tempNode->next;}return -1;}int main(){int valueInput, currentValue;int listValues;char arrow;Node* node1, * previous, * head;std::cin >> valueInput;std::cin >> currentValue;node1 = new Node();node1->value = currentValue;node1->next = NULL;previous = node1;head = node1;while (std::cin){std::cin >> arrow;std::cin >> arrow;std::cin >> currentValue;node1 = new Node();node1->value = currentValue;previous->next = node1;previous = node1;}std::cout << find(head, valueInput);}
Currently, my program returns -1 always
Sample input and outputs:
Sample Input:
5
1->2->5
Sample Output:
2
答案1
得分: 0
这是因为您的代码只有一个return -1语句。您应该使用return count而不是break。此外，您始终与0进行比较，而不是valueInput。
int find(Node* node, int valueInput){Node* tempNode = node;int count = 0;while (tempNode != nullptr){if (tempNode->value == valueInput){return count;}++count;tempNode = tempNode->next;}return -1;}
您还应该使用nullptr而不是NULL。
英文:
That's because your code only has a return -1 statement. You should return count instead of break. Also, you are always comparing against 0 instead of the valueInput
int find(Node* node, int valueInput){Node* tempNode = node;int count = 0;while (tempNode != nullptr){if (tempNode->value == valueInput){return count;}++count;tempNode = tempNode->next;}return -1;}
You should also be using nullptr instead of NULL
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