Leetcode Linked List Problem 链表问题合集

1. Leet Code OJ 2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

您将获得两个非空链接列表，表示两个非负整数。 数字以相反的顺序存储，并且它们的每个节点包含单个数字。 添加两个数字并将其作为链接列表返回。

您可以假定这两个数字不包含任何前导零，除了数字0本身。

输入：（2→4→3）+（5→6→4）

输出：7 - > 0 - > 8

代码：

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode node = new ListNode(0);
if(l1 == null && l2 == null) return l1;
back(node, l1, l2);
return node;
}

public void back(ListNode result, ListNode l1, ListNode l2){
if(l1 != null) result.val += l1.val;
else l1 = new ListNode(0);
if(l2 != null) result.val += l2.val;
else l2 = new ListNode(0);

ListNode node = new ListNode(0);
if(result.val >= 10){//说明会下一个节点值至少为1
result.val = result.val % 10;
node.val = 1;
result.next = node;
}
if( (l1.next != null || l2.next != null)){
result.next = node;
back(result.next, l1.next, l2.next);
}
}
}

2. Leet Code OJ 83. Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,

Given 1->1->2, return 1->2.

Given 1->1->2->3->3, return 1->2->3.

意思是：去除链表中重复的部分

分析

这里需要注意的点：

1. 如何返回链表

2. 当链表中存在3个以上相同的节点时的处理

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if(head == null) return null;
ListNode node = head;//1. 用一个节点来保存链头，用于返回
while(head != null){
ListNode next = head.next;
if(next != null && next.val == head.val){
head.next = next.next;
continue;//2. 当有3个以上相同的节点时，不进入下一个节点，一直到相同节点的最后那个节点时，才跳到下一个节点
}
head = head.next;
}
return node;
}
}

3. Min Stack

设计一个有最小值的Stack

分析

可以使用链表来构造

代码

public class MinStack {

private Node head;

public void push(int x) {
if (head == null) {
head = new Node(x, x);
} else {
head = new Node(x, Math.min(head.min, x), head);
}
}

public void pop() {
head = head.next;
}

public int top() {
return head.val;
}

public int getMin() {
return head.min;
}

private class Node {
int val;
int min;
Node next;

private Node(int val, int min) {
this(val, min, null);
}

private Node(int val, int min, Node next) {
this.val = val;
this.min = min;
this.next = next;
}
}
}