LeetCode 刷题记录 2020

2020.02.10

2020年计划刷 LeetCode 题目 50 道，差不多一周一道题的进度，直到今天，2020年2月10日，我还有刷一道题……

[TOC]

1. LeetCode 21, 83, 141 Easy Linked List

21. Merge Two Sorted List

​ Difficulty: Easy, Tag: Linked List

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

这个没有任何要点，算练手，不算在这50题里面，所以也就不用多说，直接上代码。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        
        ListNode head = new ListNode(0);
        ListNode p = head;
        
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                p.next = l1;
                p = p.next;
                l1 = l1.next;
            } else {
                p.next = l2;
                p = p.next;
                l2 = l2.next;
            }
        }
        
        if (l1 == null)
            p.next = l2;
        else
            p.next = l1;
        
        return head.next;
        
    }
}

83. Remove Duplicates from Sorted List

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

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

这个也没什么好说的，既然是有序的，就遍历链表，和前一个值做对比，相等就删除就可以了。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) 
            return head;
        
        int val = head.val;
        ListNode p = head;
        while (p.next != null) {
            if (p.next.val == val)
                delete(p.next, p);
            else {
                p = p.next;
                val = p.val;
            }
        }
        
        return head;
    }
    
    private void delete(ListNode node, ListNode pre) {
        pre.next = node.next;
    }
}

LeetCode 网站给出的解：

public ListNode deleteDuplicates(ListNode head) {
    ListNode current = head;
    while (current != null && current.next != null) {
        if (current.next.val == current.val) {
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }
    return head;
}

141. Linked List Cycle

Given a linked list, determine if it has a cycle in it.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

img

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

img

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

img

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

第一个思路：利用集合

遍历链表，判断当前结点是否已经存在集合中，存在的话说明有环，否则将结点加入集合。

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     LitNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null)
            return false;
        
        Set<ListNode> s = new HashSet<ListNode>();
        ListNode p = head;
        while (p != null) {
            if (s.contains(p))
                return true;
            
            s.add(p);
            p = p.next;
        }
        
        return false;
    }
}

时间复杂度：O(1), 空间复杂度: O(n)

方法二：双指针

fast(步长为2) & slow(步长为1)，如果存在环，fast 和 slow 始终不会指向空指针并且 fast 最终会和slow 指向同一个结点。

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     LitNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null)
            return false;
        
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            if (fast == slow)
                return true;
            
            slow = slow.next;
            fast = fast.next.next;
        }
        
        return false;
    }
}