83. Remove Duplicates from Sorted List (Easy)

Q:

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.

A:
 ------------------------------------------3 rd Pass ---------------------------

/**
 * 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) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        auto pre = head;
        while(pre && pre->next){
            auto cur = pre->next;
            if(cur->val == pre->val){ // remove current node
                pre->next = cur->next;
            }else{
                pre = cur;
            }
        }
        return head;
    }
};

错误：  刚开始，while内部，又用了个while，写成了

while (tail != null && tail.next != null) {
            while (tail.val == tail.next.val) {......delete the next node........}

可是这样的话，就会导致 tail.next已经指向了null指针。 



--------------------------------1st Pass -------------------------------------


就是两个指针，看是否相同。

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode p1, p2; // p1 is the current node, p2 is his next neighbor
        if(head ==null || head.next == null){
            return head;
        }
        //now we have at least 2 Nodes in the list
        p1=head;
        p2 = head.next;
        while(p2!= null){
            if(p2.val == p1.val){
                // delete p2, 
                p1.next = p2.next;
                p2 = p1.next;
            }else{
                p1= p2;
                p2 = p1.next;
            }
        }
        return head;
    }
}