LEETCODE: Remove Duplicates from Sorted List II-优快云博客

本文链接：https://blog.youkuaiyun.com/anyicheng2015/article/details/42178791

本文介绍了一种从已排序链表中移除所有具有重复数值的节点的方法，只保留原始列表中的不同数值。通过示例展示了如何将1->2->3->3->4->4->5转换为1->2->5，以及如何将1->1->1->2->3转换为2->3。

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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.

思路其实很简单，就是写起来还挺麻烦的。打完收工，明天继续刷LEETCODE！！！

/**
 * 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;
        while(head != NULL && head->next != NULL && head->val == head->next->val) {
            int curval = head->val;
            while(head->next != NULL && curval == head->next->val) {
                head = head->next;
            }
            if(head->val == curval) {
                head = head->next;
            }
        }
        
        if(head == NULL || head->next == NULL)  return head;
        ListNode *cur = head->next;
        ListNode *pre = head;
        while(cur != NULL && cur->next != NULL) {
            if(cur->val != cur->next->val) {
                pre = cur;
                cur = cur->next;
                continue;
            }
            
            int curval = cur->val;
            while(cur != NULL && cur->val == curval) {
                cur = cur->next;
            }
            pre->next = cur;
            if(cur == NULL)
                break;
        }
        
        return head;
    }
};