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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.
Example 1:
Input: 1->2->3->3->4->4->5 Output: 1->2->5
Example 2:
Input: 1->1->1->2->3 Output: 2->3
解题分析
快慢指针是个好办法！
注意推敲满指针前进的条件！
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/**
* Runtime: 1 ms, faster than 60.61% of Java online submissions for Remove Duplicates from Sorted List II.
* Memory Usage: 39.4 MB, less than 6.98% of Java online submissions for Remove Duplicates from Sorted List II.
*
* Copy from: https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/solutions/7396/kuai-man-zhi-zhen-by-powcai-2/[递归与非递归 - 删除排序链表中的重复元素 II - 力扣（LeetCode）]
*/
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode slow = dummy;
ListNode fast = dummy;
while (Objects.nonNull(fast)) {
while (Objects.nonNull(fast.next) && fast.val == fast.next.val) {
fast = fast.next;
}
if (slow.next == fast) {
slow = slow.next;
} else {
slow.next = fast.next;
}
fast = fast.next;
}
return dummy.next;
}
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/**
* 参考官方题解
*/
public ListNode deleteDuplicates(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode();
dummy.next = head;
ListNode cur = dummy;
while (cur.next != null && cur.next.next != null) {
if (cur.next.val == cur.next.next.val) {
// 将重复元素全部删除
int val = cur.next.val;
while (cur.next != null && cur.next.val == val) {
cur.next = cur.next.next;
}
} else {
cur = cur.next;
}
}
return dummy.next;
}