问题描述:
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
.
问题分析:
如题所述,给定一条升序的单链表,将链表中值重复的节点删除,只留下节点的值唯一的节点。想象维护两组指针,一组两个指针用作在原链表中遍历和操作,另一组两个指针为返回的结果链表的头指针res和对结果链表尾节点进行记录的cur指针。先遍历输入单链表,每读入一个节点,判断下一个节点值是否与当前相等,若不等,则当前节点符合返回结果的条件,将该节点加入结果链表的尾部,若相等,则继续往下比较,值相等的节点不计入返回的结果链表,此时若cur部位空,curr指向的节点尾部置为空。以此类推,遍历完输入链表后即可得到返回结果。
具体实现:
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode* pre = head;
ListNode *pos = NULL, *curr = NULL, *res = NULL;
while (pre)
{
pos = pre;
while (pre->next&&pre->next->val == pre->val)pre = pre->next;
if (pos == pre)
{
if (res == NULL)
res = pre;
else
curr->next = pre;
curr = pre;
}
else if (curr)
curr->next = NULL;
pre = pre->next;
}
return res;
}
};