题目
83. 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.
解析
class Solution_83 {
public:
ListNode *deleteDuplicates(ListNode *head) {
if (!head||!head->next)
{
return head;
}
ListNode* cur = head;
ListNode*pre = NULL;
while (cur&&cur->next)
{
pre = cur;
cur = cur->next;
ListNode* temp = pre; //记录每次重复点的开始位置
while(cur&&pre->val==cur->val)
{
pre = cur;
cur=cur->next;
}
temp->next = cur; //跳过重复位置
}
return head;
}
};
82. 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.
解析
- 由于链表开头可能会有重复项，被删掉的话头指针会改变，而最终却还需要返回链表的头指针。所以需要定义一个新的节点，然后链上原链表，然后定义一个前驱指针和一个现指针，每当前驱指针指向新建的节点，现指针从下一个位置开始往下遍历，遇到相同的则继续往下，直到遇到不同项时，把前驱指针的next指向下面那个不同的元素。如果现指针遍历的第一个元素就不相同，则把前驱指针向下移一位。
//参考容易理解一些
ListNode *deleteDuplicates(ListNode *head) {
if (!head || !head->next) return head;
ListNode *start = new ListNode(0);
start->next = head;
ListNode *pre = start;
while (pre->next) {
ListNode *cur = pre->next;
while (cur->next && cur->next->val == cur->val) cur = cur->next;
if (cur != pre->next) pre->next = cur->next;
else pre = pre->next;
}
return start->next;
}
// 82. Remove Duplicates from Sorted List II
class Solution_82 {
public:
ListNode* deleteDuplicates(ListNode* head) {
if (!head||!head->next)
{
return head;
}
ListNode*newHead = new ListNode(0);
newHead->next = head;
ListNode* pre = newHead;
ListNode* cur = head;
while (cur&&cur->next)
{
ListNode* next = cur->next;
if(next->val!=cur->val)
{
if (pre->next==cur) //pre->next当前元素开始，cur当前元素结束，cur->next另外不同的元素
{
pre = cur;
}
else
{
pre->next = cur->next;
}
}
cur = cur->next;
}
if (pre->next!=cur) //这里是地址比较，若没有重复元素，则地址相同的
{
pre->next = cur->next;
}
return newHead->next;
}
};
题目来源