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  • [LeetCode]82. Remove Duplicates from Sorted List排序链表去重

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

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

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    解法:设置两个指针curr和next指向相邻两个节点,从头往后扫描,(1)如果某次指向的两个节点值相等,则删除next指向的节点,并且next前移;(2)如果指向的两个节点值不一样,则两个节点都向前移动。
    /**
     * 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 *curr = head, *next = head->next;
            while(next != NULL) {
                if(curr->val == next->val) {
                    ListNode* del = next;
                    next = next->next;
                    curr->next = next;
                    delete del;
                }
                else {
                    curr = next;
                    next = next->next;
                }
            }
            return head;
        }
    };

    或者用一个指针:

    /**
     * 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* curr = head;
            while(curr != NULL && curr->next != NULL) {
                if(curr->val == curr->next->val) {
                    ListNode* del = curr->next;
                    curr->next = curr->next->next;
                    delete del;
                }
                else
                    curr = curr->next;
            }
            return head;
        }
    };

    需要注意的一点是可能某个重复值出现了超过2次,所以在找到重复值时不能两个指针同时前移。

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  • 原文地址:https://www.cnblogs.com/aprilcheny/p/4968055.html
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