今天小编就为大家分享一篇关于剑指offer之判断链表是否包含环，小编觉得内容挺不错的，现在分享给大家，具有很好的参考价值，需要的朋友一起跟随小编来看看吧
1 问题
判断链表是否包含环
2 思路
2个指针，一个指针走一步，一个指针走2步，如果相遇则有，反之无。
3 代码实现
#include#include #define true 1 #define false 0; typedef struct node { int value; struct node *next; }Node; /* *判断链表是否有环 */ int isCircleList(Node *head) { if (head == NULL) { return false; } Node *first = NULL; Node *second = NULL; first = head; second = head; while (second != NULL && (second->next) != NULL && (second->next->next != NULL)) { first = first->next; second = second->next->next; if (first == second) { return true; } } return false; } int main() { Node *head = NULL; Node *node1 = NULL; Node *node2 = NULL; Node *node3 = NULL; Node *node4 = NULL; Node *node5 = NULL; Node *node6 = NULL; Node *node7 = NULL; head = (Node *)malloc(sizeof(Node)); node1 = (Node *)malloc(sizeof(Node)); node2 = (Node *)malloc(sizeof(Node)); node3 = (Node *)malloc(sizeof(Node)); node4 = (Node *)malloc(sizeof(Node)); node5 = (Node *)malloc(sizeof(Node)); node6 = (Node *)malloc(sizeof(Node)); node7 = (Node *)malloc(sizeof(Node)); if (head == NULL || node1 == NULL || node2 == NULL || node3 == NULL || node4 == NULL || node5 == NULL || node6 == NULL || node7 == NULL) { printf("malloc fail\n"); return false; } // node7<-node6 <-node5 // | | //head->node1->node2->node3->node4 head->value = 0; head->next = node1; node1->value = 1; node1->next = node2; node2->value = 2; node2->next = node3; node3->value = 3; node3->next = node4; node4->value = 4; node4->next = node5; node5->value = 5; node5->next = node6; node6->value = 6; node6->next = node7; node7->value = 7; node7->next = node2; int result = isCircleList(head); if (result) { printf("list have circle\n"); } else { printf("list do not have circle\n"); } return true; }
4 运行结果
list have circle
总结
以上就是剑指offer之判断链表是否包含环的详细内容，更多请关注0133技术站其它相关文章！
