【题目】

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.

【题意】

给定一个有序链表，删除具有重复的数字，从原来的列表中只留下了不同数字的所有节点。

【分析】

思路1：

遍历链表，记录重复元素的起始位置和截止位置。要删除的重复元素的上一个位置begin，截止位置end。例如：1，2，2，2，4 begin = 0 end = 3

如果当前元素和上一个元素相等，则更新end；如果不相等则判断beigin和end是否相等，相等没有重复元素需要删除，不相等删除[being+1，end]中元素。

如果最后几个元素重复，则需要最后单独通过判断begin，end是否等来解决。

【代码1】

/*********************************
*   日期：2014-01-28
*   作者：SJF0115
*   题号: Remove Duplicates from Sorted List II
*   来源：http://oj.leetcode.com/problems/remove-duplicates-from-sorted-list-ii/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;

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;
        }
        //添加虚拟头结点
        ListNode *dummy = (ListNode*)malloc(sizeof(ListNode));
        dummy->next = head;
        //记录上一节点的值
        int preVal = head->val;
        ListNode *pre = head,*cur = head->next;
        //begin 要删除节点的上一节点  end 要删除的最后节点
        ListNode *begin = dummy,*end = dummy;
        while(cur != NULL){
            if(cur->val == preVal){
                //如果当前元素和上一个元素相等则更新删除截至元素
                end = cur;
            }
            else{
                preVal = cur->val;
                //[begin,end]有重复元素
                if(begin != end){
                    //删除重复元素
                    begin->next = end->next;
                    end = begin;
                }
                //[begin,end]没有重复元素,更新删除起始元素
                else{
                    begin = end = pre;
                }
            }
            pre = cur;
            cur = cur->next;
        }
        //如果最后几个元素相等例如：2,1,1,1,1
        if(begin != end){
            //删除重复元素
            begin->next = end->next;
        }
        return dummy->next;
    }
};
int main() {
    Solution solution;
    int A[] = {2,1,1,1,1,1};
    ListNode *head = (ListNode*)malloc(sizeof(ListNode));
    head->next = NULL;
    ListNode *node;
    ListNode *pre = head;
    for(int i = 0;i < 6;i++){
        node = (ListNode*)malloc(sizeof(ListNode));
        node->val = A[i];
        node->next = NULL;
        pre->next = node;
        pre = node;
    }
    head = solution.deleteDuplicates(head->next);
    while(head != NULL){
        printf("%d ",head->val);
        head = head->next;
    }
    return 0;
}

【代码2】

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode *dummy = (ListNode*)malloc(sizeof(ListNode));
        dummy->next = head;

        ListNode *pre = dummy,*cur = head;
        while (cur != NULL) {
            int preVal = cur->val;
            if (cur->next && cur->next->val == preVal) {
                while (cur && cur->val == preVal) {
                    pre->next = cur->next;
                    delete cur;
                    cur = pre->next;
                }
                cur = pre;
            }
            pre = cur;
            cur = cur->next;
        }
        return dummy->next;
    }
};

【代码3】

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode *dummy = (ListNode*)malloc(sizeof(ListNode));
        dummy->next = head;

        ListNode *pre = dummy,*cur = head;
        ListNode *temp;
        int i = 0;
        while (cur != NULL) {
            //判断是否有重复元素
            bool duplicated = false;
            //寻找重复元素删除
            while(cur->next != NULL && (cur->val == cur->next->val)){
                duplicated = true;
                temp = cur->next;
                //删除cur元素
                pre->next = cur->next;
                cur = temp;
            }//while
            //删除重复元素的最后一个
            if(duplicated){
                temp = cur->next;
                //删除重复元素的最后一个
                pre->next = cur->next;
                cur = temp;
                continue;
            }//if
            pre = cur;
            cur = cur->next;
        }//while
        return dummy->next;
    }
};