LeetCode-83. Remove Duplicates from Sorted List https://leetcode.com/problems/remove-duplicates-from-sorted-list/

题目描述

Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

Example 1:

Input: head = [1,1,2]
Output: [1,2]

Example 2:

Input: head = [1,1,2,3,3]
Output: [1,2,3]

Constraints:

• The number of nodes in the list is in the range [0, 300].
• -100 <= Node.val <= 100
• The list is guaranteed to be sorted in ascending order.

解题思路

 【C++解法】

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

1、递归

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head == NULL || head->next == NULL) return head;
        head->next = deleteDuplicates(head->next);
        return head->val == head->next->val ? head->next : head;
    }
};

2、循环

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode* curr = head->next;
        ListNode* prev = head;
        while (curr) {
            if (curr->val == prev->val) {prev->next = curr->next;}
            else {prev = curr;}    
            curr = prev->next;
        }
        return head;
    }
};

【Java解法】

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */

1、递归

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) {return head;}
        head.next = deleteDuplicates(head.next);
        return head.val == head.next.val ? head.next : head;
    }
}

2、循环

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) {return head;}
        ListNode pre = head;
        ListNode cur = head.next;
        while (cur != null) {
            if (pre.val == cur.val) {pre.next = cur.next;}
            else {pre = cur;}
            cur = pre.next;
        }
        return head;
    }
}