剑指offer-面试题37-两个链表的第一个公共结点

package case37_FindFirstCommonNode;import java.util.Stack;/** * 题目：输入两个链表，找出他们的第一个公共结点。 *  * @author WangSai * */public class FindFirstCommonNode {/** * 方法1： *  * 思路：从单链表的尾节点开始寻找，把两个链表分别压入栈stack1和stack2中，然后依次弹出栈，就可以实现从链表尾部开始比较。因为单链表只有一个next引用， * 所以，从第一个公共结点开始就会成Y形。从尾节点开始比较，寻找第一个不相同的地方，即可。 *  * @param head1,链表1的头结点 * @param head2，链表2的头结点 * @return 链表1和链表2的第一个公共结点。 */private static MyNode getMyFirstCommonNode(MyNode head1, MyNode head2) {// 异常值检测if (head1 == null || head2 == null)return null;// 若输入的两个结点为同一个链表的头结点if (head1 == head2)return head1;Stack<MyNode> stack1 = new Stack<>();Stack<MyNode> stack2 = new Stack<>();MyNode myCommonNode = null;// 把链表1压入栈中while (head1 != null) {stack1.push(head1);head1 = head1.next;}// 把链表2压入栈中while (head2 != null) {stack2.push(head2);head2 = head2.next;}MyNode stack1Node = stack1.pop();MyNode stack2Node = stack2.pop();while (!stack1.empty() && !stack2.empty() && stack1Node == stack2Node) {myCommonNode = stack1Node;stack1Node = stack1.pop();stack2Node = stack2.pop();}return myCommonNode;}/** * 方法2： *  * 思路：两个链表若有公共结点，则从尾部一定重合。因为单链表只有1个next引用。先分别链表1和链表2，获得两个链表的长度count1和count2，链表长度较长 * 的链表，先往前走几步Math.abs（count1-count2）,然后，一起遍历。若有公共结点，则从该结点开始，后面全部重合，呈现Y形。即可获得第一个公共结点。 *  * @param head1,链表1的头结点 * @param head2，链表2的头结点 * @return 链表1和链表2的第一个公共结点。 */private static MyNode getMyFirstCommonNode2(MyNode head1, MyNode head2) {if (head1 == null || head2 == null)return null;if (head1 == head2)return head1;MyNode n1 = head1;MyNode n2 = head2;int count1 = 0, count2 = 0;// 遍历链表1，获取元素数量while (n1 != null) {n1 = n1.next;count1++;}// 遍历链表2，获取元素数量while (n2 != null) {n2 = n2.next;count2++;}// 元素多的那个链表先开始走几步，然后，当链表剩余的元素相同时，开始比较。int dif = count2 - count1;while (dif < 0) {head1 = head1.next;dif++;}while (dif > 0) {head2 = head2.next;dif--;}while (head1 != null && head2 != null && head1 != head2) {head1 = head1.next;head2 = head2.next;}// 执行到这一步，有两种情况：// 情况1：因为两个元素head1==head2，返回head1即可。// 情况2：遍历到最后没有相同的元素。if (head1 == head2)return head1;return null;}public static void main(String[] args) {test1();System.out.println("test1()*********** over");test2();System.out.println("test2()*********** over");test3();System.out.println("test3()*********** over");test4();System.out.println("test4()*********** over");}// 链表1和链表2最后一个结点相同// 链表1：head1(0)——>N1(1)——>N2(2)——>N3(3)——>null// 链表2：head2(5)——>M1(6)——>M2(7)——>N3(3)——>nullprivate static void test1() {MyNode head1 = new MyNode(0);MyNode N1 = new MyNode(1);MyNode N2 = new MyNode(2);MyNode N3 = new MyNode(3);head1.next = N1;N1.next = N2;N2.next = N3;MyNode head2 = new MyNode(5);MyNode M1 = new MyNode(6);MyNode M2 = new MyNode(7);// MyNode M3 = new MyNode(8);head2.next = M1;M1.next = M2;M2.next = N3;System.out.println("链表1:head1(0)——>N1(1)——>N2(2)——>N3(3)——>null:");System.out.println("链表2:head2(5)——>M1(6)——>M2(7)——>N3(3)——>null:");System.out.println(getMyFirstCommonNode(head1, head2).data);System.out.println(getMyFirstCommonNode2(head1, head2).data);}// 链表1和链表2从第一个结点开始就相同,即输入同一个链表的头结点// 链表1：head1(9)——>N1(1)——>N2(2)——>N3(3)——>null// 链表2：head1(9)——>N1(1)——>N2(2)——>N3(3)——>nullprivate static void test2() {MyNode head1 = new MyNode(9);MyNode N1 = new MyNode(1);MyNode N2 = new MyNode(2);MyNode N3 = new MyNode(3);head1.next = N1;N1.next = N2;N2.next = N3;System.out.println("链表1:head1(9)——>N1(1)——>N2(2)——>N3(3)——>null:");System.out.println("链表2:head1(9)——>N1(1)——>N2(2)——>N3(3)——>null:");System.out.println(getMyFirstCommonNode(head1, head1).data);System.out.println(getMyFirstCommonNode2(head1, head1).data);}// 链表1:head1(0)——>N1(1)——>N2(2)——>N3(3)——>null// 链表2:head2(5)——>M1(6)——>M2(7)——>M3(8)——>null// 没有公共结点private static void test3() {MyNode head1 = new MyNode(0);MyNode N1 = new MyNode(1);MyNode N2 = new MyNode(2);MyNode N3 = new MyNode(3);head1.next = N1;N1.next = N2;N2.next = N3;MyNode head2 = new MyNode(5);MyNode M1 = new MyNode(6);MyNode M2 = new MyNode(7);MyNode M3 = new MyNode(8);head2.next = M1;M1.next = M2;M2.next = M3;System.out.println("链表1:head1(0)——>N1(1)——>N2(2)——>N3(3)——>null:");System.out.println("链表2:head2(5)——>M1(6)——>M2(7)——>M3(8)——>null:");System.out.println(getMyFirstCommonNode(head1, head2));System.out.println(getMyFirstCommonNode2(head1, head2));}// 链表1:head1(0)——>N1(1)——>N2(2)——>N3(3)——>null// 链表2:head2(5)——>M1(6)——>N2(2)——>N3(3)——>null// 从中部开始相同。private static void test4() {MyNode head1 = new MyNode(0);MyNode N1 = new MyNode(1);MyNode N2 = new MyNode(2);MyNode N3 = new MyNode(3);head1.next = N1;N1.next = N2;N2.next = N3;MyNode head2 = new MyNode(5);MyNode M1 = new MyNode(6);// MyNode M2 = new MyNode(7);// MyNode M3 = new MyNode(8);head2.next = M1;M1.next = N2;System.out.println("链表1:head1(0)——>N1(1)——>N2(2)——>N3(3)——>null:");System.out.println("链表2:head2(5)——>M1(6)——>N2(2)——>N3(3)——>null:");System.out.println(getMyFirstCommonNode(head1, head2).data);System.out.println(getMyFirstCommonNode2(head1, head2).data);}}