leetcode解题报告（6）：Remove Duplicates from Sorted List

描述

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

只需要从头结点开始遍历，判断当前结点元素是否和下一个元素相等，如果相等，则跳过下一结点，否则直接让当前结点等于下一结点。

代码如下：

/**
 * 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* tmp = head;
        while(tmp){
            while(tmp -> next && tmp -> val == tmp -> next -> val)
                tmp -> next = tmp -> next -> next;
            tmp = tmp -> next;

        }
        return head;
    }
};

另一解法如下：

https://discuss.leetcode.com/topic/33179/recommend-for-beginners-c-implementation-with-detailed-explaination

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head == NULL)    return NULL;
        ListNode* pre = head,*cur = head -> next;
        while(cur){
            if(pre -> val == cur -> val){
                pre -> next = cur -> next;
                cur = cur -> next;
            }else{
                pre -> val = cur -> val;
                pre = pre -> next;
                cur = cur -> next;
            }
        }
};