【LeetCode】Remove Duplicates from Sorted List

题目描述

Remove Duplicates from Sorted List

 

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.

题目分析

去除链表中的重复数字。纯粹的链表操作。

定义两个指针，如果后面那个指针的val和前面的一样，就把后面的删除

代码示例

#include <iostream>#include <vector>using namespace std;/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */  struct ListNode {      int val;      ListNode *next;      ListNode(int x) : val(x), next(NULL) {} };  void printList(ListNode *head,char *name){printf("%s\n",name);while(head != NULL){printf("%3d,",head->val);head = head->next;}printf("\n");}class Solution {public:    ListNode *deleteDuplicates(ListNode *head) {        if(head == NULL)return NULL;        ListNode *p = head;        ListNode *q = p->next;        while(q != NULL)        {        if(p->val == q->val)//把q移除{p->next = q->next;q = p->next; } else{q = q->next;p = p->next;}}return head;    }};/*vector 在添加过程中会动态的申请内存，元素的位置会动态的调整，所以不能边创建vector边创建链表，中间指针可能会动。这里的list需要使用传引用 */ListNode* listCreate(vector<ListNode> &list, int arr[], int n){if(n <= 0)return NULL;for(int i=0;i<n;i++){ListNode tmp(arr[i]);list.push_back(tmp);}ListNode *head = &list[0];for(int i=1;i<n;i++){head->next = &list[i];head = head->next;}head = &list[0];return head;}void CheckList(ListNode *head,int arr[], int n){int i = 0;for(i = 0;i<n && head != NULL;i++,head=head->next){if(head->val != arr[i])break;}if(i != n)printf("------------------------failed\n");elseprintf("------------------------passed\n");}//功能测试，长度为偶数 void test0(){vector<ListNode> list;int arr[10] = {0,1,1,3,3,5,6,7,7,7};ListNode *head = listCreate(list,arr, 10);printList(head,"head");Solution so;so.deleteDuplicates(head);int arr1[6] = {0,1,3,5,6,7};CheckList(head,arr1,6);//printList(head,"head");}int main(int argc, char *argv[]){test0(); return 0;}

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