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定义:
循环链表:链表中一个节点的next指针指向先前已经存在的节点,导致链表中出现环。
问题1:判断是否有环
#include <cstring>
#include <iostream>using namespace std;struct node
{char value;node* next;node(char rhs){value = rhs;next = NULL;}
};bool isLoop(node* head)
{if (head == NULL){return false;}node* slow = head;node* fast = head;while((fast!= NULL) && (fast->next != NULL)){slow = slow->next;fast = fast->next->next;if (slow == fast){break;}}return !(fast == NULL || fast->next == NULL);
}int main()
{node A('A');node B('B');node C('C');node D('D');node E('E');node F('F');node G('G');node H('H');node I('I');node J('J');node K('K');A.next = &B;B.next = &C;C.next = &D;D.next = &E;E.next = &F;F.next = &G;G.next = &H;H.next = &I;I.next = &J;J.next = &K;K.next = &D;if (isLoop(&A)){cout<<"Loop";}else{cout<<"No loop";}return 0;
}
问题2:找到这个环的起始点
输入: A->B->C->D->E->F->G->H->I->J->K->D
输出:D
分析:
当fast与slow相遇时, slow肯定没有遍历完链表,而fast在环内肯定循环了1圈以上。
设环的长度为r, 相遇时fast在环内走了n个整圈(n > 1),slow走了s步,fast走了2s步,则:
2s = s + nr -> s = nr
设整个链表的长度为L,环入口点与相遇点的距离为x,链表起点到环入口点的距离为a,则:
a + x = s = nr (slow走过的步数,slow为走过一圈)
a + x = (n-1)r + r = (n-1)r + (L - a) -> a = (n-1)r + (r - x)
(r - x) 为相遇点到环入口点的距离; 因此,链表头到环入口点的距离 等于 (n-1)个环循环 + 相遇点到环入口的距离。
我们从链表头和相遇点分别设置一个指针,每次各走一步,则两个指针必定相遇,且第一个相遇点为环入口点。
#include <cstring>
#include <iostream>using namespace std;struct node
{char value;node* next;node(char rhs){value = rhs;next = NULL;}
};node* isLoop(node* head)
{if (head == NULL){return false;}node* slow = head;node* fast = head;while((fast!= NULL) && (fast->next != NULL)){slow = slow->next;fast = fast->next->next;if (slow == fast){break;}}if (fast == NULL || fast->next == NULL){return NULL;}// currently, the list is loopedslow = head;while(slow != fast){slow = slow->next;fast = fast->next;}return slow;
}int main()
{node A('A');node B('B');node C('C');node D('D');node E('E');node F('F');node G('G');node H('H');node I('I');node J('J');node K('K');A.next = &B;B.next = &C;C.next = &D;D.next = &E;E.next = &F;F.next = &G;G.next = &H;H.next = &I;I.next = &J;J.next = &K;K.next = &D;node* p;if ((p= isLoop(&A))!= NULL){cout<<"Loop, the interaction node is "<<p->value;}else{cout<<"No loop";}return 0;
}
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