Practice47:
在一个排序的链表中,存在重复的结点,请删除该链表中重复的结点,重复的结点不保留,返回链表头指针。 例如,链表1->2->3->3->4->4->5 处理后为 1->2->5
S1
牛客暴力解法
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* deleteDuplication(ListNode* pHead)
{
if(!pHead)return NULL;
set<int> st;
ListNode *pre=pHead;
ListNode *cur=pHead->next;
while(cur){
if(pre->val==cur->val){
st.insert(pre->val);
}
pre = pre->next;
cur = cur->next;
}
ListNode *vhead = new ListNode(-1);
vhead->next = pHead;
pre = vhead;
cur = pHead;
while(cur){
if(st.count(cur->val)){
cur = cur->next;
pre->next = cur;
}
else{
cur = cur->next;
pre = pre->next;
}
}
return vhead->next;
}
};
S2
S1优化版本
class Solution {
public:
ListNode* deleteDuplication(ListNode* pHead)
{
ListNode *vhead = new ListNode(-1);
vhead->next = pHead;
ListNode *pre = vhead, *cur = pHead;
while (cur) {
if (cur->next && cur->val == cur->next->val) {
cur = cur->next;
while (cur->next && cur->val == cur->next->val) {
cur = cur->next;
}
cur = cur->next;
pre->next = cur;
}
else {
pre = cur;
cur = cur->next;
}
}
return vhead->next;
}
};