每日一题之删除排序链表重复元素

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描述：

给一个排好序的链表，如 1->2->3->3->5; 删除其中的重复元素之后结果为 1->2->5

思路：

遍历链表，用pre指针记录当前节点的前一个节点，然后每次遍历的时候比较 p->data 与 p->next->data。其中需要注意的是，需要一个标记，来处理pre是直接后移，还是需要再跳过重复元素。

#include <bits/stdc++.h>

using namespace std;

struct Node {

	int data;
	Node* next;
	Node(int x):data(x),next(nullptr){}
};

Node* deletSameElement(Node* List) {

	Node* p = List;
	Node* pre = p;
	Node* res = List;
	p = p->next;
	bool flag = 0;
	while(p != nullptr) {
		if (p->next != nullptr) {
			if (p->data == p->next->data) {
				p = p->next;
				flag = 1;
			}
			else {
				if (flag == 0) {
					pre = p;
					p = p->next;
				}
				else {
					pre->next = p->next;
					p = p->next;
					flag = 0;
				}
			}
		}
		else break;
	}

	return res;
}

void build() {
	Node* List = new Node(0); 
	Node* tmp = new Node(1);
	List->next = tmp;
	Node* tmp2 = new Node(2);
	tmp->next = tmp2;
	Node* tmp3 = new Node(3);
	tmp2->next = tmp3;
	Node* tmp4 = new Node(3);
	tmp3->next = tmp4;
	Node* tmp5 = new Node(5);
	tmp4->next = tmp5;

	Node* p = List;
	while(p != nullptr) {
		cout << p->data << " ";
		p = p->next;
	}

	cout << endl;

	Node* res = deletSameElement(List);

	while(res != nullptr) {
		cout << res->data <<" ";
		res = res->next;
	}
}

int main() {

	build();

	return 0;
}