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.
思路： Easy. 依第二个开始，从前往后走一遍，若下一个相同，则删掉，迈过去，否则，走一步。
/**
* 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 *p = head;
while(p && p->next) {
if(p->val == p->next->val) {
ListNode *q = p->next;
p->next = p->next->next;
free(q);
}
else p = p->next;
}
return head;
}
};
Remove Duplicates from Sorted List II
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.
思路，由于前面几个相同时，比较麻烦，难以统一处理。故在头结点前面再加一个伪头结点（值小于头结点），这样前后走一遍即可。（代码未释放空间）
/**
* 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 *pseudoHead = new ListNode(0);
pseudoHead->next = head;
ListNode *pre, *cur;
pre = pseudoHead;
while(head) {
for(cur = head->next; cur && cur->val == head->val; cur = cur->next);
if(head->next != cur) {
pre->next = cur;
head = cur;
}
else { pre = head; head = head->next; }
}
return pseudoHead->next;
}
};
或
/**
* 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 pseudoHead(0);
pseudoHead.next = head;
ListNode *pre, *cur;
pre = &pseudoHead;
while(head) {
for(cur = head->next; cur && cur->val == head->val; cur = cur->next);
if(head->next != cur) {
pre->next = cur;
head = cur;
}
else { pre = head; head = head->next; }
}
return pseudoHead.next;
}
};
