若鏈表中存在環(huán)，找出其中的環(huán)所在的節(jié)點，否則，返回NULL

在沒有C++ set容器的優(yōu)勢前提下，我們對這樣的環(huán)型的尋找以及定位可以利用快慢指針來實現(xiàn)。
有環(huán)的存在，類似與操場跑圈，必然存在快慢之分。有了快慢，就一定會有交點。反之，有了交點，就一定存在環(huán)。

實現(xiàn)過程如下：

1. slow指針移動一次，fast指針移動兩次，當fast指針和slow指針相等時保留相等時的節(jié)點
2. 根據(jù)運算，從相等時的節(jié)點和頭節(jié)點以相同的速度運算，當兩者相等時即為環(huán)的起始節(jié)點

算法實現(xiàn)如下：

Data *get_circle_list(Data *head) {Data *fast;Data *slow;fast  = head;slow = head;Data *meet = NULL;/*快速和慢速指針一起移動*/while (fast) {fast = fast -> next;slow = slow -> next;if (!fast) {//當快速指針為空時，即不存在環(huán)型節(jié)點return NULL;}//否則快速指針繼續(xù)移動fast = fast -> next;if(fast == slow) {meet = fast;break;}}/*如果沒有找到相遇的節(jié)點，那么即不存在環(huán)型*/if(meet == NULL) {return NULL;} else {while (!(head == meet)) {head = head -> next;meet = meet -> next;}return head;}return NULL;
}

源代碼測試如下:

#include <stdio.h>
#include <stdlib.h>typedef struct link_list {int data;struct link_list *next;
}Data;void print_list(Data *head) {while (head) {printf("%d ",head->data);head = head -> next;}printf("\n");
}Data *insert_head(Data *head, int n) {head = (Data *)malloc (sizeof(Data));head -> next = NULL;int data;while(n --) {Data *p = (Data *)malloc(sizeof(Data));scanf("%d",&data);p -> data = data;p -> next = head -> next;head -> next = p;}return head;
}Data *insert_tail(Data *head, int n) {head = (Data *)malloc(sizeof(Data));head -> next = NULL;Data *r = head;int data;while(n--) {Data *p = (Data *)malloc(sizeof(Data));scanf("%d", &data);p -> data = data;r -> next = p;r = p;p -> next = NULL;}return head;
}Data *get_circle_list(Data *head) {Data *fast;Data *slow;fast  = head;slow = head;Data *meet = NULL;while (fast) {fast = fast -> next;slow = slow -> next;if (!fast) {return NULL;}fast = fast -> next;if(fast == slow) {meet = fast;break;}}if(meet == NULL) {return NULL;} else {while (!(head == meet)) {head = head -> next;meet = meet -> next;}return head;}
}int main() {/*構(gòu)造環(huán)形測試用例*/Data a1,a2,a3,a4,a5,a6,a7;a1.data = 1;a2.data = 2;a3.data = 3;a4.data = 4;a5.data = 5;a6.data = 6;a7.data = 7;a1.next = &a2;a2.next = &a3;a3.next = &a4;a4.next = &a5;a5.next = &a6;a6.next = &a7;/*構(gòu)造環(huán)形，這里很明顯環(huán)形的起點為5節(jié)點*/a7.next = &a5;printf("get circle node \n");Data *result = get_circle_list(&a1);printf("get circle node is %d\n",result -> data);return 0;
}

輸出如下:

get circle node 
get circle node is 5

總結(jié)

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