Category | Difficulty | Likes | Dislikes |
---|
algorithms | Medium (45.93%) | 221 | - |
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给定一个排序链表,删除所有含有重复数字的节点,只保留原始链表中 *没有重复出现 *的数字。
示例 1:
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2
| 输入: 1->2->3->3->4->4->5
输出: 1->2->5
|
示例 2:
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2
| 输入: 1->1->1->2->3
输出: 2->3
|
Discussion | Solution
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| impl Solution {
pub fn delete_duplicates(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
let mut dummy_head = Some(Box::new(ListNode::new(0)));
let mut ptr = &mut dummy_head;
let mut pre_val: i32 = -1;
let mut pre_val_init = false;
let mut next_val: i32 = 0;
let mut head = head;
while let Some(mut node) = head {
head = node.next.take();
if head.is_some() {
if node.val == head.as_ref().unwrap().val {
pre_val_init = true;
pre_val = node.val;
continue
}
}
if pre_val_init && pre_val == node.val {
continue
}
pre_val = node.val;
pre_val_init = true;
ptr.as_mut().unwrap().next = Some(node);
ptr = &mut ptr.as_mut().unwrap().next;
}
dummy_head.unwrap().next
}
}
|