Divided into three cases:
the first case: a linked list has a ring, the ring is not a list, it is unlikely that two lists intersect
the second case: two lists do not ring
the third case: there are two lists ring
public class FindFirstIntersectNode { public static class Node { public int value; public Node next; public Node(int data) { this.value = data; } } /** * Divided into three cases: * The first case: a linked list has a ring, the ring is not a list, it is impossible that two lists intersect. * The second case: two lists did not ring, * The third case: two lists are rings, */ public static Node getIntersectNode(Node head1, Node head2) { if (head1 == null || head2 == null) { return null; } Node loop1 = getLoopNode(head1); Node loop2 = getLoopNode(head2); // two chain acyclic intersects IF (Loop1 == null && Loop2 == null ) { return noloop (head1, head2 The); } // two looped chain intersects IF (Loop1 =! Null && Loop2 =! Null ) { return bothLoop (head1, Loop1, head2 The, Loop2); } return null; } /** * Determining whether a chain ring, a chain looped back to the first node of the ring, the ring without or null * @param head * @return * / Public static the Node getLoopNode (the Node head) { // first determined speed of the pointer meeting point cross the Node = FAST head; Node slow = head; Node cross = null; while (fast.next != null && fast.next.next != null) { fast = fast.next.next; slow = slow.next; if (slow == fast) { cross = fast; break; } } // starting from the meeting point and the head of the list, each time a mobile node, they met the entrance point is ring the Node Start = head; the while (! = Start null ! && Cross = null ) { IF (Start == Cross) { return Cross; } start = start.next; cross = cross.next; } return null; } /** * Two chain acyclic intersect, the intersection returns the first intersection node, or null disjoint * @param head1 * @Param head2 * @return */ public static Node noLoop(Node head1, Node head2) { if (head1 == null || head2 == null) { return null; } Node cur1 = head1; int len1 = 1; Node cur2 = head2; int len2 = 1; while(cur1.next != null) { len1 ++ ; cur1 = cur1.next; } while(cur2.next != null) { len2++; cur2 = cur2.next; } // the last node is not the same, two disjoint lists IF (Cur1 =! CUR2) { return null ; } // First always points to the beginning of a long chain, second chain always points to the beginning of the short, diff represents the difference in length of two lists the Node First = null ; SECOND the Node = null ; int the diff = 0 ; IF (LEN1> LEN2) { // list 2 lists longer than 1, First = head1; second = head2; diff = len1 -len2; }else { first = head2; second = head1; diff = len2 - len1; } // premise diff is not zero, go a long chain diff the while (diff> 0 ) { first = first.next; diff--; } // First and second take Meanwhile, if the same, showing the addition of the while (First! = Second) { first =first.next; second = second.next; } return first; } /** * There are two chain rings intersect, the intersection returns the first intersection node, or null disjoint * @Param head1 node of the first head of the list * @Param loop1 first list of ingress node * @Param head2 second linked list head node * @Param loop2 ingress point of the second list * @return */ public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) { Node cur1 = null; CUR2 the Node = null ; IF (== Loop1 Loop2) { // corresponds to a case where two and, the list is determined two acyclic intersect the same procedure, except the end node into a Loop1 Cur1 = head1; cur2 = head2; int n = 0; while (cur1 != loop1) { n++; cur1 = cur1.next; } while (cur2 != loop2) { n--; cur2 = cur2.next; } cur1 = n > 0 ? head1 : head2; cur2 = cur1 == head1 ? head2 : head1; n = Math.abs(n); while (n != 0) { n--; cur1 = cur1.next; } while (cur1 != cur2) { cur1 = cur1.next; cur2 = cur2.next; } return cur1; } The else { // corresponding to the case where three and four, in which case a three-phase, the case of a four disjoint Cur1 = loop1.next; the while (! = Cur1 Loop1) { IF (Cur1 == Loop2) { return Loop1; } cur1 = cur1.next; } return null; } } public static void main(String[] args) { // 1->2->3->4->5->6->7->null Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(4); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(6); head1.next.next.next.next.next.next = new Node(7); // 0->9->8->6->7->null Node head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next.next.next.next.next; // 8->6 System.out.println(getIntersectNode(head1, head2).value); // 1->2->3->4->5->6->7->4... head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(4); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(6); head1.next.next.next.next.next.next = new Node(7); head1.next.next.next.next.next.next.next = head1.next.next.next; // 7->4 // 0->9->8->2... head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next; // 8->2 System.out.println(getIntersectNode(head1, head2).value); // 0->9->8->6->4->5->6.. head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next.next.next.next.next; // 8->6 System.out.println(getIntersectNode(head1, head2).value); } }
A ring intersects two-way linked list example of FIG.
Acyclic two-way linked list example of FIG intersect
