两个单链表相交的一系列问题------牛客网初级班练习14
【题目】 在本题中,单链表可能有环,也可能无环。给定两个 单链表的头节点 head1和head2,这两个链表可能相交,也可能 不相交。请实现一个函数, 如果两个链表相交,请返回相交的 第一个节点;如果不相交,返回null 即可。 要求:如果链表1 的长度为N,链表2的长度为M,时间复杂度请达到 O(N+M),额外 空间复杂度请达到O(1)。
1)如何判断一个单链表有环还是无环
方法一:
遍历链表元素,放入HashSet集合中
方法二:
准备俩个指针,一个快指针,一个慢指针,快指针一次走俩步,慢指针一次走一步
如果有环,快指针和慢指针一定会相遇
当出现相遇情况,快指针回到起始节点,并且由一次走俩步变成一次走一步
快指针和慢指针一定会在第一个入环节点相遇
2)如何判断俩个无环链表第一个相交节点
方法一:
将遍历第一个链表,放入map中,遍历第二个链表,判断map中是否有这个元素
方法二:
遍历链表一,统计链表一的长度len1以及或者链表一最后一个节点end1
遍历链表二,统计链表二的长度len2以及或者链表二最后一个节点end2
比较end1和end2内存地址是否相等,如果end1不等于end2,则不可能相交
如果相等,比较len1和len2长度,假如len1=100,len2=80,链表一先走20步。。。。
如果一个链表有环,一个链表无环,不可能相交
3)如何判断俩个有环链表第一个相交节点
1.俩个环不相交
2.先相交后共享一个环
3.
如图:
准备工作:
head1,loop1
head2,loop2
如果loop1等于loop2,属于结构2
如果loop1不等于loop2,可能是结构1或者结构3
代码如下:
public static class Node {
public int value;
public Node next;
public Node(int data) {
this.value = data;
}
}
public static Node getIntersectNode(Node head1, Node head2) {
if (head1 == null || head2 == null) {
return null;
}
Node loop1 = getLoopNode(head1);
Node loop2 = getLoopNode(head2);
if (loop1 == null && loop2 == null) {
return noLoop(head1, head2);
}
if (loop1 != null && loop2 != null) {
return bothLoop(head1, loop1, head2, loop2);
}
return null;
}
public static Node getLoopNode(Node head) {
if (head == null || head.next == null || head.next.next == null) {
return null;
}
Node n1 = head.next; // n1 -> slow
Node n2 = head.next.next; // n2 -> fast
while (n1 != n2) {
if (n2.next == null || n2.next.next == null) {
return null;
}
n2 = n2.next.next;
n1 = n1.next;
}
n2 = head; // n2 -> walk again from head
while (n1 != n2) {
n1 = n1.next;
n2 = n2.next;
}
return n1;
}
public static Node noLoop(Node head1, Node head2) {
if (head1 == null || head2 == null) {
return null;
}
Node cur1 = head1;
Node cur2 = head2;
int n = 0;
while (cur1.next != null) {
n++;
cur1 = cur1.next;
}
while (cur2.next != null) {
n--;
cur2 = cur2.next;
}
if (cur1 != cur2) {
return null;
}
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;
}
public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
Node cur1 = null;
Node cur2 = null;
if (loop1 == loop2) {
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;
} else {
cur1 = loop1.next;
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 = 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);
}