leecode 解题总结：83. Remove Duplicates from Sorted List

#include <iostream>#include <stdio.h>#include <vector>using namespace std;/*问题: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.分析:删除链表中的冗余元素，多个元素只保留一个。我们保留多个重复元素中最后一个节点。同时设定当前节点和下一个节点。如果当前节点和下一个节点相同，说明当前元素是重复的，记录当前元素之前的节点previoudNode，然后一直寻找到和当前元素不同的节点，记为nextNode,则nextNode前面一个节点就是需要保留的节点node，然后设定previouNode指向node即可输入:3(结点个数)1 1 24 1 1 2 22 1 151 1 2 3 3输出:1 21 211 2 3关键:1】 1,1,2,2报错了//删除完之后，进行连接if(previous){previous->next = node;}//删除头部节点，则重新设置链表头部else{head = node;}//这里需要更新previous为当前node(保留的重复元素的最后一个节点)，否则会出现错误，记住previous记录的就是与下一个元素不重复的节点previous = node;*/struct ListNode {int val;ListNode *next;ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode* deleteDuplicates(ListNode* head) {        if(!head){return head;}//只有一个元素直接返回if(!(head->next)){return head;}ListNode* previous = NULL;ListNode* node = head;ListNode* next = head->next;ListNode* tempNode;ListNode* deleteNode;int repeatValue;int countRepeat = 0;while(next){//判断当前元素和下一个元素是否重复if(node->val == next->val){//寻找第一个不重复的元素repeatValue = node->val;countRepeat = 0;while(next && next->val == repeatValue ){node = next;next = next->next;countRepeat++;}//此时node是最后一个重复的元素，删除重复元素if(previous){deleteNode = previous->next;}//说明链表头部就重复了else{deleteNode = head;}while(deleteNode && countRepeat){tempNode = deleteNode->next;delete deleteNode;deleteNode = tempNode;countRepeat--;}//删除完之后，进行连接if(previous){previous->next = node;}//删除头部节点，则重新设置链表头部else{head = node;}//这里需要更新previous为当前node(保留的重复元素的最后一个节点)，否则会出现错误previous = node;}else{previous = node;}node = next;if(next){next = next->next;}}return head;    }};void print(ListNode* head){if(!head){cout << "no result" << endl;}ListNode* tempHead = head;while(head){cout << head->val << " ";head = head->next;}cout << endl;head = tempHead;}ListNode* buildList(vector<int>& nums){if(nums.empty()){return NULL;}int size = nums.size();ListNode* head ;ListNode *tail;ListNode* node;for(int i = 0 ; i < size ; i++){if(i){node = new ListNode(nums.at(i));tail->next = node;tail = node;}else{head = new ListNode(nums.at(i));tail = head;}}return head;}void deleteList(ListNode* head){ListNode* node;while(head){node = head->next;delete head;head = node;}}void process(){ vector<int> nums; int value; int num; Solution solution; vector<int> result; while(cin >> num ) { nums.clear(); for(int i = 0 ; i < num ; i++) { cin >> value; nums.push_back(value); } ListNode* head = buildList(nums); ListNode* newHead = solution.deleteDuplicates(head); print(newHead); deleteList(newHead);//删除节点了 }}int main(int argc , char* argv[]){process();getchar();return 0;}