C指针插入算法

我创建了两个结构：

typedef struct mainNode {
    int theNode;
    int visited;
    struct mainNode *next;
} node_rec, *node_ptr;

及

我需要将其插入列表中：

void insert(node_ptr head, int theNodeVal, int neighborNodeVal, int weight) {

//if list is empty
if (head == NULL) {
    head = (node_ptr) malloc(sizeof (node_rec)); //create head of list
    head->theNode = theNodeVal; //set head value to node value
    head->next = NULL; //point to null       
}

//while list is not pointing no null
while (head != NULL) {
    //if node IS NOT equal to node value   
    if (head->theNode != theNodeVal) {
        head->theNode = theNodeVal; //set head value (new node) to node value
        head->next = tail; //connect to next node
        tail->next = NULL; //point to null
    }
    else
    {          
        //if node IS equal to node value (the node already exists)
        tail->next = head // head is the new tail
        head->neighbor = n; //point at the neighbor of head (new tail)
    }
}
}

我想看看我实现的逻辑是否正确。这就是我评论每一行的原因。您可以参考页面顶部的链接以获得视觉效果。

将

if（head->theNode=theNodeVal）{

替换为

if（head->theNode==theNodeVal）{

。这会有帮助吗

这是我将如何实现该算法的图形：

#include <limits.h>
#include <string.h>
#include <malloc.h>
#include <stdio.h>

struct Vertex {
    struct Vertex *next; // The reference to the next vertex in the global list of vertices
    int nodeId;
    int distance;
    struct Vertex *previous; // The reference to trace back the shortest path;
    struct Edge *neighbor; // The reference to the list of edges of this 
};

struct Edge {
    struct Edge *next;   // The reference to the next edge of this vertex;
    struct Vertex *node; // The reference to the vertex at other end of this edge;
    int weight;
};


struct Vertex *head = NULL;

struct Vertex *newVertex(int nodeId) {
    struct Vertex *p = malloc(sizeof(struct Vertex));
    memset(p,0,sizeof(struct Vertex));
    p->nodeId = nodeId;
    p->distance=INT_MAX;
    p->next = head;
    head = p;
    return p;
}

void addEdge(struct Vertex *v1, struct Vertex *v2, int weight) {
    struct Edge *e = malloc(sizeof(struct Edge));
    e->next = v1->neighbor;
    v1->neighbor = e;
    e->node = v2;
    e->weight = weight;
}

void insert(int node1, int node2, int weight) {
    struct Vertex *current = head, *p1 = NULL, *p2 = NULL;

    while (current != NULL) {
        if (current->nodeId == node1) p1 = current;
        if (current->nodeId == node2) p2 = current;
        current = current->next;
    }

    if (p1 == NULL) p1 = newVertex(node1);
    if (p2 == NULL) p2 = newVertex(node2);

    addEdge(p1,p2,weight);
    addEdge(p2,p1,weight);
}

void main(int argc, char **argv) {
    int node1, node2, weight;
    while (EOF != scanf("%d %d %d \n", &node1, &node2, &weight)) {
        insert(node1, node2, weight);
    }
}

#包括
#包括
#包括
#包括
结构顶点{
struct Vertex*next；//对全局顶点列表中下一个顶点的引用
int-nodeId；
整数距离；
struct Vertex*previous；//追溯最短路径的引用；
struct Edge*neighbor；//对该对象的边列表的引用
};
结构边{
struct Edge*next；//对该顶点下一条边的引用；
struct Vertex*node；//对该边另一端顶点的引用；
整数权重；
};
结构顶点*head=NULL；
结构顶点*新顶点（int nodeId）{
结构顶点*p=malloc（sizeof（结构顶点））；
memset（p，0，sizeof（struct-Vertex））；
p->nodeId=nodeId；
p->distance=INT_MAX；
p->next=头部；
水头=p；
返回p；
}
void addEdge（结构顶点*v1，结构顶点*v2，整数权重）{
结构边缘*e=malloc（sizeof（结构边缘））；
e->next=v1->邻居；
v1->邻居=e；
e->node=v2；
e->重量=重量；
}
无效插入（int节点1、int节点2、int权重）{
结构顶点*current=head，*p1=NULL，*p2=NULL；
while（当前！=NULL）{
如果（当前->节点ID==节点1）p1=当前；
如果（当前->节点ID==节点2）p2=当前；
当前=当前->下一步；
}
如果（p1==NULL）p1=newVertex（node1）；
如果（p2==NULL）p2=newVertex（node2）；
附加值（p1、p2、重量）；
附加值（p2、p1、重量）；
}
void main（整型argc，字符**argv）{
int节点1、节点2、权重；
while（EOF！=scanf（“%d%d%d\n”、&node1、&node2、&weight））{
插入（节点1、节点2、重量）；
}
}

问题是什么？你在坚持什么？至于逻辑，我想

struct main node

的成员

neighbor

应该声明为neighbor\u node\ptr。if两个

语句的主体非常相似。我建议将其排除在普通事物之外。这样，它将更具可读性，更清晰再次遇到同样的问题。
=
是一个作业！我会稍微修改你的帖子+1我发誓我需要编写一个正则表达式，扫描一个文件，查看控制流语句表达式，然后进行一次符号运算，而不是相等性测试。我只能想象，用这样一只野兽回答一些SO问题的速度有多快。眼光真好。
#include <limits.h>
#include <string.h>
#include <malloc.h>
#include <stdio.h>

struct Vertex {
    struct Vertex *next; // The reference to the next vertex in the global list of vertices
    int nodeId;
    int distance;
    struct Vertex *previous; // The reference to trace back the shortest path;
    struct Edge *neighbor; // The reference to the list of edges of this 
};

struct Edge {
    struct Edge *next;   // The reference to the next edge of this vertex;
    struct Vertex *node; // The reference to the vertex at other end of this edge;
    int weight;
};


struct Vertex *head = NULL;

struct Vertex *newVertex(int nodeId) {
    struct Vertex *p = malloc(sizeof(struct Vertex));
    memset(p,0,sizeof(struct Vertex));
    p->nodeId = nodeId;
    p->distance=INT_MAX;
    p->next = head;
    head = p;
    return p;
}

void addEdge(struct Vertex *v1, struct Vertex *v2, int weight) {
    struct Edge *e = malloc(sizeof(struct Edge));
    e->next = v1->neighbor;
    v1->neighbor = e;
    e->node = v2;
    e->weight = weight;
}

void insert(int node1, int node2, int weight) {
    struct Vertex *current = head, *p1 = NULL, *p2 = NULL;

    while (current != NULL) {
        if (current->nodeId == node1) p1 = current;
        if (current->nodeId == node2) p2 = current;
        current = current->next;
    }

    if (p1 == NULL) p1 = newVertex(node1);
    if (p2 == NULL) p2 = newVertex(node2);

    addEdge(p1,p2,weight);
    addEdge(p2,p1,weight);
}

void main(int argc, char **argv) {
    int node1, node2, weight;
    while (EOF != scanf("%d %d %d \n", &node1, &node2, &weight)) {
        insert(node1, node2, weight);
    }
}