leetcode(83)—— 刪除已排序連結串列重複元素(Remove Duplicates from Sorted List)
思路:維護兩指標(ListNode *p1, ListNode *p2,p2 始終指向 p1 的next),分別指向當前節點值和下一個節點,如果當前節點的值等於下一個節點的值,則…,如果不相等,則…。
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if (!head) return NULL;
ListNode *p1 = head, *p2 = head->next;
while (p1 && p2)
{
if (p1->val == p2->val)
{
p1->next = p2->next;
p2 = p1->next;
// p2 始終指向 p1 的 next
}
else
{
p1 = p2;
p2 = p1->next;
// p2 始終指向 p1 的 next
}
}
return head;
}
};
上述實現可能略顯囉嗦:
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode *cur = head;
while (cur && cur->next)
{
if (cur->val == cur->next->val)
cur-> next = cur->next->next;
else
cur = cur->next;
}
return head;
}
};
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