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package byteDance;

/**

* Created by lizeyang on 2020/5/12.

* 两个链表，可能相交，找出相交的节点

*/

class ListNode {

int val;

ListNode next;

ListNode(int val) {

this.val = val;

}

}

public class test07 {

//无辅助空间，时间复杂度O(n)

public static int test(ListNode root1, ListNode root2) {

if (root1 == null || root2 == null) {

return 0;

}

int count = 0;

ListNode p = root1;

ListNode q = root2;

while (p != null) {

count++;

p = p.next;

}

while (q != null) {

count--;

q = q.next;

}

ListNode head1 = root1;

ListNode head2 = root2;

if (count > 0) {

while (count > 0) {

head1 = head1.next;

count--;

}

} else {

while (count != 0) {

head2 = head2.next;

count++;

}

}

ListNode h1 = root1;

ListNode h2 = root2;

while (h1 != null && h2 != null && h1 != h2) {

h1 = h1.next;

h2 = h2.next;

}

return h1.val;

}

public static void main(String[] args) {

ListNode head1 = new ListNode(1);

head1.next = new ListNode(2);

head1.next.next = new ListNode(3);

ListNode head2 = new ListNode(4);

head2.next = head1.next.next;

System.out.println(test(head1, head2));

}

}