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224 lines
5.8 KiB
224 lines
5.8 KiB
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链表相关
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---
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[06. 从头到尾打印链表](https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/)
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```java
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class Solution {
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public int[s] reversePrint(ListNode head) {
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ListNode temp = head;
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int size = 0;
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while (temp != null) {
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temp = temp.next;
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size++;
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}
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int[] result = new int[size];
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int index = size - 1;
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while (head != null) {
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result[index--] = head.val;
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head = head.next;
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}
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return result;
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}
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}
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```
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```java
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class Solution {
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public int[] reversePrint(ListNode head) {
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Stack<ListNode> stack = new Stack<ListNode>();
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ListNode temp = head;
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while (temp != null) {
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stack.push(temp);
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temp = temp.next;
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}
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int size = stack.size();
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int[] print = new int[size];
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for (int i = 0; i < size; i++) {
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print[i] = stack.pop().val;
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}
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return print;
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}
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}
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```
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[22. 链表中倒数第 k 个节点](https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/)
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```java
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class Solution {
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public ListNode getKthFromEnd(ListNode head, int k) {
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int length = 0;
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ListNode temp = head;
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while (temp != null) {
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temp = temp.next;
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length++;
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}
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for (int i = 0; i < length - k; i++) {
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head = head.next;
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}
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return head;
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}
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}
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```
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```java
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class Solution {
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public ListNode getKthFromEnd(ListNode head, int k) {
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ListNode h0 = head;
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for (int i = 0; i < k; i++) {
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h0 = h0.next;
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}
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while (h0 != null) {
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h0 = h0.next;
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head = head.next;
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}
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return head;
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}
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}
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```
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[24. 反转链表](https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/)
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```java
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class Solution {
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public ListNode reverseList(ListNode head) {
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if (head == null || head.next == null) {
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return head;
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}
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ListNode h1 = head;
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ListNode h2 = head.next;
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ListNode h3 = null;
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h1.next = null;
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while (h2 != null) {
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h3 = h2.next;
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h2.next = h1;
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h1 = h2;
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h2 = h3;
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}
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return h1;
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}
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}
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```
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```java
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class Solution {
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public ListNode reverseList(ListNode head) {
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// 递归终止条件是当前为空,或者下一个节点为空
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if (head == null || head.next == null) {
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return head;
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}
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// 这里的 h1 就是最后一个节点
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ListNode h1 = reverseList(head.next);
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// 如果链表是 1->2->3->4->5,那么此时的 cur 就是 5
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// 而 head 是4,head的 下一个是 5,下下一个是空
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// 所以 head.next.next 就是 5->4
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head.next.next = head;
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// 防止链表循环,需要将 head.next 设置为空
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head.next = null;
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// 每层递归函数都返回 h1,也就是最后一个节点
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return h1;
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}
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}
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```
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[25. 合并两个排序的链表](https://leetcode-cn.com/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof/)
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```java
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class Solution {
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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
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ListNode h0 = new ListNode(0);
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ListNode h = h0;
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while (l1 != null && l2 != null) {
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if (l1.val <= l2.val) {
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h0.next = l1;
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l1 = l1.next;
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} else {
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h0.next = l2;
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l2 = l2.next;
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}
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h0 = h0.next;
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}
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if (l1 == null) {
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h0.next = l2;
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} else {
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h0.next = l1;
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}
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return h.next;
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}
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}
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```
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```java
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class Solution {
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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
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if (l1 == null) {
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return l2;
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}
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if (l2 == null) {
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return l1;
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}
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if (l1.val <= l2.val) {
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l1.next = mergeTwoLists(l1.next, l2);
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return l1;
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} else {
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l2.next = mergeTwoLists(l1, l2.next);
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return l2;
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}
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}
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}
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```
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[35. 复杂链表的复制](https://leetcode-cn.com/problems/fu-za-lian-biao-de-fu-zhi-lcof/)
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```java
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class Solution {
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public Node copyRandomList(Node head) {
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// 存原节点和拷贝节点的一个映射
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Map<Node, Node> map = new HashMap<>();
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Node temp = head;
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while (temp != null) {
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map.put(temp, new Node(temp.val));
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temp = temp.next;
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}
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temp = head;
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while (temp != null) {
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// 将拷贝节点组织成一个链表
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map.get(temp).random = map.get(temp.random);
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map.get(temp).next = map.get(temp.next);
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temp = temp.next;
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}
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return map.get(head);
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}
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}
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```
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```java
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class Solution {
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public Node copyRandomList(Node head) {
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// 将拷贝节点放在原节点后面,比如 1->2 变成 1->1'->2->2'
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for (Node node = head, copy = null; node != null; node = node.next.next) {
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copy = new Node(node.val);
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copy.next = node.next;
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node.next = copy;
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}
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// 把拷贝节点的 random 指针安排上
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for (Node node = head; node != null; node = node.next.next) {
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if (node.random != null) {
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node.next.random = node.random.next;
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}
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}
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Node newHead = head.next;
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Node node = head;
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Node temp = null;
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// 分离原节点和拷贝节点
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while (node != null && node.next != null) {
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temp = node.next;
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node.next = temp.next;
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node = temp;
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}
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return newHead;
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}
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}
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```
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