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;
        }
    }
}