82. Remove Duplicates from Sorted List II【Medium】

82. Remove Duplicates from Sorted List II【Medium】

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.

解法一：

/**
 * 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 == NULL || head->next == NULL) {
            return head;
        }
        
        ListNode * dummy = new ListNode(INT_MIN);
        dummy->next = head;
        head = dummy;
        int duplicate;
        
        while (head->next != NULL && head->next->next != NULL) {
            if (head->next->val == head->next->next->val) {
                duplicate = head->next->val;
                while (head->next != NULL && head->next->val == duplicate) {
                    ListNode * temp = head->next;
                    free(temp);
                    head->next = head->next->next;
                }
            }
            else {
                head = head->next;
            }
        }
        return dummy->next;
        
    }
};

nc

解法二：

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(!head) return head;
        ListNode dummy(0);
        dummy.next = head;
        ListNode *pre = &dummy;
        ListNode *cur = head;

        while(cur && cur->next){
            while(cur->next && cur->val == cur->next->val) cur = cur->next;
            if(pre->next == cur){
                pre = cur;
                cur = cur->next;
            } else {
                cur = cur->next;
                pre->next = cur;
            }
        }

        return dummy.next;
    }
};

参考了@liismn 的代码

解法三：

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(!head||!head->next) return head;
        ListNode* dummy = new ListNode(0);
        ListNode* tail = dummy;
        int flag = true; // should the current head be added ?
        while(head){
            while(head&&head->next&&head->val==head->next->val)
            {
                flag = false; // finds duplicate, set it to false
                head = head->next;
            }
            if(flag) // if should be added
            {
                tail->next = head;
                tail = tail->next;
            }
            head = head->next;
            flag = true; // time for a new head value, set flag back to true
        }
        tail->next = nullptr; // Don't forget this... I did..
        return dummy->next;
    }
};

参考了@GoGoDong 的代码

解法四：

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

递归，参考了@totalheap 的代码，还不太明白