1 问题
输入两个链表，找出它们的第一个公共结点。
含有公共节点的两个链表的结构类似于下图中的链表：
1 -> 2 -> 3 -> 4 ->5
2 -> 4 ->5
可以看到两个链表中有一个公共节点，其中4节点就是这两个链表的公共节点
2 分析
既然题目是求公共节点，说明一定存在这个节点，然后我们可以发现两个链表的尾巴是一样，都重合了是Y性结构，我们先把长的链表的头移动到短的头那里，然后一个接着下一个比较就行
3 代码实现
#include <stdio.h>
#include <stdlib.h>typedef struct Node
{int value;struct Node *next;
} Node;/** 初始化结构体*/
struct Node* init(struct Node *node, int value)
{node = (struct Node*)malloc(sizeof(Node));if (node != NULL){node->value = value;//这个地方不要忘记设置为NULLnode->next = NULL;return node;}return NULL;
}/** 获取链表的长度*/
int length(Node *head)
{if (head == NULL)return 0;Node *p = head;int length = 0;while (p != NULL){length++;p = p->next;}return length;
}/*** 找到第一个公共的节点*/
struct Node* get_common(Node *head1, Node *head2)
{if (head1 == NULL || head2 == NULL){return NULL;}int list1_length = length(head1);int list2_length = length(head2);Node *short_head = NULL;Node *long_head = NULL;int sub_len = 0;if (list1_length > list2_length){short_head = head2;long_head = head1;sub_len = list1_length - list2_length;}else{short_head = head1;long_head = head2;sub_len = list2_length - list1_length;}//移动长链表，确保两个链表一样长while (sub_len > 0){sub_len--;long_head = long_head->next;}while (short_head != NULL && long_head != NULL){if (short_head->value == long_head->value){return short_head;}short_head = short_head->next;long_head = long_head->next;}return NULL;
}int main()
{Node *n1 = NULL;Node *n2 = NULL;Node *n3 = NULL;Node *n4 = NULL;Node *n5 = NULL;Node *m1 = NULL;Node *m2 = NULL;Node *m3 = NULL;n1 = init(n1, 1);n2 = init(n2, 2);n3 = init(n3, 3);n4 = init(n4, 4);n5 = init(n5, 5);m1 = init(m1, 2);m2 = init(m2, 4);m3 = init(m3, 5);if (n1 && n2 && n3 && n4 && n5){n1->next = n2;n2->next = n3;n3->next = n4;n4->next = n5;}if (m1 && m2 && m3){m1->next = m2;m2->next = m3;}Node *node = get_common(n1, m2);if (node){printf("common node value is: %d\n", node->value);}else{printf("two list do not common value\n");}if (n1) {free(n1); n1 = NULL;}if (n2) {free(n2); n2 = NULL;}if (n3) {free(n3); n3 = NULL;}if (n4) {free(n4); n4 = NULL;}if (n5) {free(n5); n5 = NULL;}if (m1) {free(m1); m1 = NULL;}if (m2) {free(m2); m1 = NULL;}if (m3) {free(m3); m1 = NULL;}return 1;
}
4 运行结果
common node value is: 4
5 总结
如果我们求链表的长度，一般是这样的函数
/** 获取链表的长度*/
int length(Node *head)
{if (head == NULL)return 0;Node *p = head;int length = 0;while (p != NULL){length++;p = p->next;}return length;
}一定要记到骨髓里面去。
