第二章 链表问题
2.1 打印两个有序链表的公共部分
【题目】
给定两个有序链表的头指针 head1 和 head2,打印两个链表的公共部分。
【难度】
士 ★☆☆☆
【题解】
给定两个链表士有序链表,可以从两个链表的头开始进行如下判断:
- 如果 head1 的值小于 head2 的值,则 head1 往下移动;
- 如果 head2 的值小于 head1 的值,则 head2 往下移动;
- 如果 head1 和 head2 的值相等,则打印此值,head1 与 head2 都往下移动;
- 直到 head1 或 head2 任何一个移动到 null,整个过程结束。
【实现】
- TwoLinkedListCommonPart.java
import java.util.ArrayList;
import java.util.List;
public class TwoLinkedListCommonPart {
private static class Node {
public int value;
public Node next;
public Node(int value) {
this.value = value;
}
}
private Node head1;
private Node head2;
private List<Integer> common;
public TwoLinkedListCommonPart(int[] arr1, int[] arr2) {
build(arr1, arr2);
}
private void build(int[] arr1, int[] arr2) {
if (arr1 == null || arr1.length == 0) {
this.head1 = null;
} else {
this.head1 = new Node(arr1[0]);
Node head = head1;
for (int i = 1; i < arr1.length; ++i) {
head = head.next = new Node(arr1[i]);
}
}
if (arr2 == null || arr2.length == 0) {
this.head2 = null;
} else {
this.head2 = new Node(arr2[0]);
Node head = head2;
for (int i = 1; i < arr2.length; ++i) {
head = head.next = new Node(arr2[i]);
}
}
}
public List<Integer> getCommon() {
if (this.common == null) {
calculate();
}
return this.common;
}
private void calculate() {
this.common = new ArrayList<>();
Node head1 = this.head1;
Node head2 = this.head2;
while (head1 != null && head2 != null) {
if (head1.value > head2.value) {
head1 = head1.next;
} else if (head1.value < head2.value) {
head2 = head2.next;
} else {
common.add(head1.value);
head1 = head1.next;
head2 = head2.next;
}
}
}
}
- TwoLinkedListCommonPartTest.java
import java.util.List;
public class TwoLinkedListCommonPartTest {
public static void main(String[] args) {
int[] arr1 = {6, 5, 4, 3, 2};
int[] arr2 = {4, 3, 2, 1, 0};
TwoLinkedListCommonPart instance = new TwoLinkedListCommonPart(arr1, arr2);
List<Integer> common = instance.getCommon();
}
}