[leetcode] 13. Remove Duplicates from Sorted List

这个题目其实不难的,主要是我C++的水平太差了,链表那里绊了好久,但是又不像用python,所以还是强行上了。

题目如下:

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.

思路很简单,拿到链表后依次遍历,遇到next->val == val;就把next的节点删掉就行。然后就看C++的操作了。

题解如下:

/**
 * 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) 
    {
        ListNode *aspros = head;
        ListNode *deferos = NULL;

        if (aspros != NULL)
        {
            while (aspros->next)
            {
                int nextVal = aspros->next->val;
                while (aspros->val != nextVal && aspros->next->next)
                {
                    deferos = aspros;
                    aspros = aspros->next;
                    nextVal = aspros->next->val;
                }
                if (aspros->val == nextVal)
                {
                    if (deferos == NULL)
                    {
                        head = aspros->next;
                    }
                    else
                    {
                        deferos->next = aspros->next;

                    }

                }
                else
                {

                    // 这里是正常链表删完所有重复节点后的出口
                    return head;
                }
                aspros = aspros->next;
            }

            // 这里处理只给入一个值的链表的情况
            return head;
        }
        else
        {
            // 这里处理给入空链表的情况
            return head;
        }
    }
};
posted @ 2014-11-22 16:37  Tiny-Box  阅读(128)  评论(0编辑  收藏  举报