linear-list/linked-list/remove-duplicates-from-sorted-list-ii

Remove Duplicates from Sorted List II

描述

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,

Given 1->2->3->3->4->4->5, return 1->2->5.

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

分析

无

递归版

// Remove Duplicates from Sorted List II
// 递归版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head ==null || head.next == null) return head;

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

// Remove Duplicates from Sorted List II
// 递归版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        if (head == nullptr || head->next == nullptr) return head;

        ListNode *p = head->next;
        if (head->val == p->val) {
            while (p != nullptr && head->val == p->val) {
                ListNode *tmp = p;
                p = p->next;
                delete tmp;
            }
            delete head;
            return deleteDuplicates(p);
        } else {
            head->next = deleteDuplicates(head->next);
            return head;
        }
    }
};

迭代版

// Remove Duplicates from Sorted List II
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) return head;

        ListNode dummy = new ListNode(Integer.MAX_VALUE); // 头结点
        dummy.next = head;
        ListNode prev = dummy, cur = head;
        while (cur != null) {
            boolean duplicated = false;
            while (cur.next != null && cur.val == cur.next.val) {
                duplicated = true;
                cur = cur.next;
            }
            if (duplicated) { // 删除重复的最后一个元素
                cur = cur.next;
                continue;
            }
            prev.next = cur;
            prev = prev.next;
            cur = cur.next;
        }
        prev.next = cur;
        return dummy.next;
    }
}

// Remove Duplicates from Sorted List II
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        if (head == nullptr) return head;

        ListNode dummy(INT_MIN); // 头结点
        dummy.next = head;
        ListNode *prev = &dummy, *cur = head;
        while (cur != nullptr) {
            bool duplicated = false;
            while (cur->next != nullptr && cur->val == cur->next->val) {
                duplicated = true;
                ListNode *temp = cur;
                cur = cur->next;
                delete temp;
            }
            if (duplicated) { // 删除重复的最后一个元素
                ListNode *temp = cur;
                cur = cur->next;
                delete temp;
                continue;
            }
            prev->next = cur;
            prev = prev->next;
            cur = cur->next;
        }
        prev->next = cur;
        return dummy.next;
    }
};

相关题目

• Remove Duplicates from Sorted List