Remove Duplicates from Sorted List & II

Remove Duplicates from Sorted List

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

Ex1

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

Ex2

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

Iterative Version

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == nullptr) return head;

        ListNode* prev = head;
        ListNode* cur = head->next;
        // for (; cur; cur = cur->next) { //bug
        for (; cur; cur = prev->next) {
            if (prev->val == cur->val) { // delete node, do not move prev pointer
                prev->next = cur->next;
                delete cur;
            } else { // advance prev
                //prev = cur;  // also correct
                prev = prev->next;
            }
        }
        return head;
    }
};

Recursive Version

// Remove Duplicates from Sorted List
// 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) return head;
        ListNode dummy(head->val + 1); // 值只要跟head不同即可
        dummy.next = head;

        recur(&dummy, head);
        return dummy.next;
    }
private:
    static void recur(ListNode *prev, ListNode *cur) {
        if (cur == nullptr) return;

        if (prev->val == cur->val) { // 删除head
            prev->next = cur->next;
            delete cur;
            recur(prev, prev->next);
        } else {
            recur(prev->next, cur->next);
        }
    }
};

Remove Duplicates from Sorted List II

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

Ex1

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Ex2

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

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == nullptr) return head;

        ListNode dummy(-1);
        ListNode *prev = &dummy;
        prev->next = head;
        ListNode *cur = head;
        ListNode *ad = head->next;
        if (ad == nullptr) return head;

        while (ad) {
            if (cur->val != ad->val) { 
                prev = cur;
                cur = ad;
                ad = ad->next;
            } else { // delete both cur and ad 1: not tail, 2, tail case
                while (ad != nullptr && cur->val == ad->val) {
                    delete cur; // bug case when input is [1,1], since we delete head
                    cur = ad;
                    ad = ad->next;
                }
                delete cur;
                prev->next = ad;
                if (ad == nullptr) {
                    break;
                } else {
                    cur = ad;
                    ad = cur->next;  
                }
            }
        }
        return dummy.next;
    }
};

bug case

input:[1,1]
output: []

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == nullptr) return head;

        ListNode dummy(-1);
        ListNode *prev = &dummy;
        prev->next = head;
        ListNode *cur = head;
        ListNode *ad = head->next;
        if (ad == nullptr) return head;

        while (ad) {
            if (cur->val != ad->val) { 
                prev = cur;
                cur = ad;
                ad = ad->next;
            } else { // delete both cur and ad 1: not tail, 2, tail case
                while (cur->val == ad->val && ad != nullptr) {
                    delete cur;  // bug case when input is [1,1], since we delete head
                    cur = ad;
                    ad = ad->next;
                }
                if (ad == nullptr) {
                    delete cur; // bug case when input is [1,1], since we delete head
                    break;
                } else {
                    prev->next = ad;
                    delete cur;
                    cur = ad;
                    ad = cur->next;  
                }
            }
        }
        return dummy.next;
    }
};

Other Solution

// 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;
    }
};

// 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;
        }
    }
};