355 lines
10 KiB
Java
355 lines
10 KiB
Java
<!-- GFM-TOC -->
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* [1. 找出两个链表的交点](#1-找出两个链表的交点)
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* [2. 链表反转](#2-链表反转)
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* [3. 归并两个有序的链表](#3-归并两个有序的链表)
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* [4. 从有序链表中删除重复节点](#4-从有序链表中删除重复节点)
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* [5. 删除链表的倒数第 n 个节点](#5-删除链表的倒数第-n-个节点)
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* [6. 交换链表中的相邻结点](#6-交换链表中的相邻结点)
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* [7. 链表求和](#7-链表求和)
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* [8. 回文链表](#8-回文链表)
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* [9. 分隔链表](#9-分隔链表)
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* [10. 链表元素按奇偶聚集](#10-链表元素按奇偶聚集)
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<!-- GFM-TOC -->
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链表是空节点，或者有一个值和一个指向下一个链表的指针，因此很多链表问题可以用递归来处理。
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# 1. 找出两个链表的交点
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[160. Intersection of Two Linked Lists (Easy)](https://leetcode.com/problems/intersection-of-two-linked-lists/description/)
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例如以下示例中 A 和 B 两个链表相交于 c1：
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```html
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A: a1 → a2
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↘
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c1 → c2 → c3
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↗
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B: b1 → b2 → b3
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```
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但是不会出现以下相交的情况，因为每个节点只有一个 next 指针，也就只能有一个后继节点，而以下示例中节点 c 有两个后继节点。
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```html
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A: a1 → a2 d1 → d2
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↘ ↗
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c
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↗ ↘
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B: b1 → b2 → b3 e1 → e2
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```
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要求时间复杂度为 O(N)，空间复杂度为 O(1)。如果不存在交点则返回 null。
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设 A 的长度为 a + c，B 的长度为 b + c，其中 c 为尾部公共部分长度，可知 a + c + b = b + c + a。
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当访问 A 链表的指针访问到链表尾部时，令它从链表 B 的头部开始访问链表 B；同样地，当访问 B 链表的指针访问到链表尾部时，令它从链表 A 的头部开始访问链表 A。这样就能控制访问 A 和 B 两个链表的指针能同时访问到交点。
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如果不存在交点，那么 a + b = b + a，以下实现代码中 l1 和 l2 会同时为 null，从而退出循环。
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```java
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public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
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ListNode l1 = headA, l2 = headB;
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while (l1 != l2) {
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l1 = (l1 == null) ? headB : l1.next;
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l2 = (l2 == null) ? headA : l2.next;
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}
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return l1;
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}
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```
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如果只是判断是否存在交点，那么就是另一个问题，即 [编程之美 3.6]() 的问题。有两种解法：
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- 把第一个链表的结尾连接到第二个链表的开头，看第二个链表是否存在环；
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- 或者直接比较两个链表的最后一个节点是否相同。
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# 2. 链表反转
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[206. Reverse Linked List (Easy)](https://leetcode.com/problems/reverse-linked-list/description/)
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递归
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```java
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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 next = head.next;
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ListNode newHead = reverseList(next);
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next.next = head;
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head.next = null;
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return newHead;
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}
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```
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头插法
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```java
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public ListNode reverseList(ListNode head) {
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ListNode newHead = new ListNode(-1);
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while (head != null) {
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ListNode next = head.next;
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head.next = newHead.next;
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newHead.next = head;
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head = next;
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}
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return newHead.next;
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}
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```
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# 3. 归并两个有序的链表
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[21. Merge Two Sorted Lists (Easy)](https://leetcode.com/problems/merge-two-sorted-lists/description/)
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```java
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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
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if (l1 == null) return l2;
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if (l2 == null) return l1;
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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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# 4. 从有序链表中删除重复节点
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[83. Remove Duplicates from Sorted List (Easy)](https://leetcode.com/problems/remove-duplicates-from-sorted-list/description/)
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```html
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Given 1->1->2, return 1->2.
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Given 1->1->2->3->3, return 1->2->3.
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```
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```java
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public ListNode deleteDuplicates(ListNode head) {
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if (head == null || head.next == null) return head;
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head.next = deleteDuplicates(head.next);
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return head.val == head.next.val ? head.next : head;
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}
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```
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# 5. 删除链表的倒数第 n 个节点
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[19. Remove Nth Node From End of List (Medium)](https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/)
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```html
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Given linked list: 1->2->3->4->5, and n = 2.
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After removing the second node from the end, the linked list becomes 1->2->3->5.
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```
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```java
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public ListNode removeNthFromEnd(ListNode head, int n) {
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ListNode fast = head;
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while (n-- > 0) {
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fast = fast.next;
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}
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if (fast == null) return head.next;
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ListNode slow = head;
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while (fast.next != null) {
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fast = fast.next;
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slow = slow.next;
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}
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slow.next = slow.next.next;
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return head;
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}
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```
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# 6. 交换链表中的相邻结点
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[24. Swap Nodes in Pairs (Medium)](https://leetcode.com/problems/swap-nodes-in-pairs/description/)
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```html
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Given 1->2->3->4, you should return the list as 2->1->4->3.
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```
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题目要求：不能修改结点的 val 值，O(1) 空间复杂度。
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```java
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public ListNode swapPairs(ListNode head) {
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ListNode node = new ListNode(-1);
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node.next = head;
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ListNode pre = node;
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while (pre.next != null && pre.next.next != null) {
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ListNode l1 = pre.next, l2 = pre.next.next;
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ListNode next = l2.next;
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l1.next = next;
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l2.next = l1;
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pre.next = l2;
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pre = l1;
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}
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return node.next;
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}
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```
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# 7. 链表求和
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[445. Add Two Numbers II (Medium)](https://leetcode.com/problems/add-two-numbers-ii/description/)
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```html
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Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
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Output: 7 -> 8 -> 0 -> 7
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```
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题目要求：不能修改原始链表。
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```java
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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
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Stack<Integer> l1Stack = buildStack(l1);
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Stack<Integer> l2Stack = buildStack(l2);
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ListNode head = new ListNode(-1);
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int carry = 0;
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while (!l1Stack.isEmpty() || !l2Stack.isEmpty() || carry != 0) {
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int x = l1Stack.isEmpty() ? 0 : l1Stack.pop();
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int y = l2Stack.isEmpty() ? 0 : l2Stack.pop();
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int sum = x + y + carry;
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ListNode node = new ListNode(sum % 10);
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node.next = head.next;
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head.next = node;
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carry = sum / 10;
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}
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return head.next;
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}
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private Stack<Integer> buildStack(ListNode l) {
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Stack<Integer> stack = new Stack<>();
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while (l != null) {
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stack.push(l.val);
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l = l.next;
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}
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return stack;
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}
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```
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# 8. 回文链表
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[234. Palindrome Linked List (Easy)](https://leetcode.com/problems/palindrome-linked-list/description/)
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题目要求：以 O(1) 的空间复杂度来求解。
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切成两半，把后半段反转，然后比较两半是否相等。
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```java
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public boolean isPalindrome(ListNode head) {
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if (head == null || head.next == null) return true;
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ListNode slow = head, fast = head.next;
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while (fast != null && fast.next != null) {
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slow = slow.next;
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fast = fast.next.next;
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}
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if (fast != null) slow = slow.next; // 偶数节点，让 slow 指向下一个节点
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cut(head, slow); // 切成两个链表
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return isEqual(head, reverse(slow));
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}
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private void cut(ListNode head, ListNode cutNode) {
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while (head.next != cutNode) {
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head = head.next;
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}
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head.next = null;
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}
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private ListNode reverse(ListNode head) {
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ListNode newHead = null;
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while (head != null) {
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ListNode nextNode = head.next;
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head.next = newHead;
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newHead = head;
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head = nextNode;
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}
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return newHead;
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}
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private boolean isEqual(ListNode l1, ListNode l2) {
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while (l1 != null && l2 != null) {
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if (l1.val != l2.val) return false;
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l1 = l1.next;
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l2 = l2.next;
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}
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return true;
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}
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```
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# 9. 分隔链表
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[725. Split Linked List in Parts(Medium)](https://leetcode.com/problems/split-linked-list-in-parts/description/)
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```html
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Input:
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root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
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Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
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Explanation:
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The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
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```
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题目描述：把链表分隔成 k 部分，每部分的长度都应该尽可能相同，排在前面的长度应该大于等于后面的。
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```java
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public ListNode[] splitListToParts(ListNode root, int k) {
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int N = 0;
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ListNode cur = root;
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while (cur != null) {
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N++;
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cur = cur.next;
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}
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int mod = N % k;
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int size = N / k;
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ListNode[] ret = new ListNode[k];
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cur = root;
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for (int i = 0; cur != null && i < k; i++) {
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ret[i] = cur;
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int curSize = size + (mod-- > 0 ? 1 : 0);
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for (int j = 0; j < curSize - 1; j++) {
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cur = cur.next;
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}
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ListNode next = cur.next;
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cur.next = null;
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cur = next;
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}
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return ret;
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}
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```
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# 10. 链表元素按奇偶聚集
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[328. Odd Even Linked List (Medium)](https://leetcode.com/problems/odd-even-linked-list/description/)
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```html
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Example:
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Given 1->2->3->4->5->NULL,
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return 1->3->5->2->4->NULL.
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```
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```java
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public ListNode oddEvenList(ListNode head) {
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if (head == null) {
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return head;
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}
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ListNode odd = head, even = head.next, evenHead = even;
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while (even != null && even.next != null) {
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odd.next = odd.next.next;
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odd = odd.next;
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even.next = even.next.next;
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even = even.next;
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}
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odd.next = evenHead;
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return head;
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}
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```
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# 微信公众号
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更多精彩内容将发布在微信公众号 CyC2018 上，你也可以在公众号后台和我交流学习和求职相关的问题。另外，公众号提供了该项目的 PDF 等离线阅读版本，后台回复 "下载" 即可领取。公众号也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
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