作业:链表交叉点

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考尔菲德
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发布于: 2020 年 07 月 29 日

解题思路:

如果两个链表存在交叉点的话,图形如下:

意味着在交叉点之后,两个链表的节点是一样的。这就意味着如果两个链表不一样长的话,长的链表多出来的节点,针对上图就是指d肯定不可能是交叉点,所以,解题方案就是跳过长链表多出来的部分,从e开始与短链表从a开始,逐个节点进行比较,如果节点相同即为交叉节点,具体的代码如下:

构建节点对象,代码如下:

static class Node {
private int value;
private Node next;
public Node(int value) {
this.value = value;
}
public int getValue() {
return value;
}
public Node getNext() {
return next;
}
public void setNext(Node next) {
this.next = next;
}
@Override
public String toString() {
return "Node{" +
"value=" + value +
'}';
}
}



求两个链表的交叉点,方法如下:

public static Node findIntersection(Node head1, Node head2) {
int size1 = sizeOf(head1);
int size2 = sizeOf(head2);
if (size1 == 0 || size2 == 0) {
return null;
}
Node longHead = null;
Node shortHead = null;
int offset = size1 - size2;
if (offset >= 0) {
longHead = head1;
shortHead = head2;
} else {
longHead = head2;
shortHead = head1;
offset = -offset;
}
Node lcNode = longHead;
Node scNode = shortHead;
if (offset > 0) {
int count = 1;
lcNode = lcNode.getNext();
while (offset > count) {
lcNode = lcNode.getNext();
count++;
}
}
while (lcNode != null) {
if (lcNode.getValue() == scNode.getValue()) {
return lcNode;
}
lcNode = lcNode.getNext();
scNode = scNode.getNext();
}
return null;
}
private static int sizeOf(Node head) {
if (head == null) {
return 0;
}
int size = 1;
Node start = head;
while (start.getNext() != null) {
size++;
start = start.getNext();
}
return size;
}

客户端代码如下:

public static void main(String[] args) {
Node node1 = new Node(1);
Node node2 = new Node(2);
Node node3 = new Node(3);
Node node4 = new Node(4);
Node node5 = new Node(5);
Node node6 = new Node(6);
Node node7 = new Node(7);
Node node8 = new Node(8);
Node node9 = new Node(9);
node1.setNext(node2);
node2.setNext(node3);
node3.setNext(node4);
node4.setNext(node5);
node5.setNext(node6);
node6.setNext(node7);
node7.setNext(node8);
node8.setNext(node9);
Node node11 = new Node(11);
Node node22 = new Node(22);
Node node33 = new Node(33);
node11.setNext(node22);
node22.setNext(node33);
node33.setNext(node6);
Node intersection = findIntersection(node11, node1);
System.out.println(intersection);
}

最后的打印输出结果如下:

Node{value=6}



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作业:链表交叉点