Remove Duplicates from Sorted Lists II @leetcode

刷题必备书籍：Cracking the Coding Interview: 150 Programming Questions and Solutions 

简历：The Google Resume: How to Prepare for a Career and Land a Job at Apple, Microsoft, Google, or any Top Tech Company
算法学习书籍：Introduction to Algorithms
编程珠玑：Programming Pearls (2nd Edition)
C++ 学习：The C++ Programming Language, 4th Edition
经典操作系统书籍，龙书：Operating System Concepts
创业：The Start-up of You: Adapt to the Future, Invest in Yourself, and Transform Your Career

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.

这道题做了好久逻辑还是没对，总是漏掉一些东西，考虑得很复杂，看了别人的代码后有了一些启发。第一，用safeguard，就是在head前面加上一个头结点，避免处理head的情况。最后记得删掉。第二，就是平时可以先用自己复杂的逻辑把代码写出来，再对比别人的看如何把逻辑简化，这样就可以把代码简化，多做几次会形成一种直觉，下一次设计会快很多。

	/**
	* 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(head == NULL) return head;
	ListNode *G = new ListNode(INT_MIN);
	G->next = head;
	ListNode *cur = G, *pre = head;
	while(pre!=NULL)
	{
	bool isDup = false;
	while(pre->next!=NULL && pre->val == pre->next->val)
	{
	isDup = true;
	pre = pre->next;
	}
	if(isDup)
	{
	pre = pre->next;
	continue;
	}
	cur->next = pre;
	cur = cur->next;
	pre= pre->next;
	}
	cur->next = pre;
	ListNode *temp = G->next;
	delete G;
	return temp;
	}
	};

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