[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。

转载于:https://www.cnblogs.com/aprilcheny/p/4968055.html