演算法詳解 有環鏈表__演算法詳解

定義：

迴圈鏈表：鏈表中一個節點的next指標指向先前已經存在的節點，導致鏈表中出現環。

問題1：判斷是否有環

#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;}

問題2：找到這個環的起始點

輸入： A->B->C->D->E->F->G->H->I->J->K->D

輸出：D

分析：

當fast與slow相遇時， slow肯定沒有遍曆完鏈表，而fast在環內肯定迴圈了1圈以上。

設環的長度為r, 相遇時fast在環內走了n個整圈(n > 1)，slow走了s步，fast走了2s步，則：

2s = s + nr    ->  s = nr

設整個鏈表的長度為L，環進入點與相遇點的距離為x，鏈表起點到環進入點的距離為a,則：

a + x = s = nr    (slow走過的步數，slow為走過一圈)

a + x = (n-1)r + r = (n-1)r + (L - a)    ->  a = (n-1)r + (r - x)

(r - x) 為相遇點到環進入點的距離； 因此，鏈表頭到環進入點的距離 等於 (n-1)個環迴圈 + 相遇點到環入口的距離。

我們從鏈表頭和相遇點分別設定一個指標，每次各走一步，則兩個指標必定相遇，且第一個相遇點為環進入點。

#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 looped   slow = 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;}