[面试真题] LeetCode：Remove Duplicates from Sorted List I & II

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.

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(NULL == head){
            return head;
        }
        int dup = head->val;
        ListNode *p = head;
        while(p->next){
            if(dup == p->next->val){
                p->next = p->next->next;
            }else{
                dup = p->next->val;
                p = p->next;
            }
        }
        return head;
    }
};

Run Status: Accepted!
Program Runtime: 84 milli secs

Progress: 164/164 test cases passed.

 

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.

解法： 记录重复部分左相邻节点tail，isdup用于判断当前节点是否重复。鉴于输入可能从一开始就重复，需设置新的头节点left。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(NULL == head){
             return head;
         }
         int dup = head->val;
         ListNode *left = new ListNode(0);
         left->next = head;
         ListNode *tail = left, *p = head;
         bool isdup = false;
         while(p->next){
             if(dup == p->next->val){
                 p = p->next;
                 tail->next = p->next;
                 isdup = true;
             }else{
                 if(!isdup){
                     tail = p;
                 }
                 dup = p->next->val;
                 p = p->next;
                 isdup = false;
             }
         }
         return left->next;
    }
};

Run Status: Accepted!
Program Runtime: 60 milli secs

Progress: 166/166 test cases passed.