LeetCode Remove Duplicates from Sorted List & Remove Duplicates from Sorted List II

Remove Duplicates from Sorted List

 

Description:

Given a sorted linked list, delete all duplicates such that each element appear only once.

Solution:

每次while循环，都判断当前节点与下一个节点的val是否一样，一样则改变当前节点的next指针，否则跳到下一个节点。

import java.util.*;public class Solution {public ListNode deleteDuplicates(ListNode head) {ListNode temp = head;ListNode next;while (temp != null) {next = temp.next;if (next == null)break;if (temp.val == next.val) {temp.next = next.next;} else {temp = next;}}return head;}}

Remove Duplicates from Sorted List II

 Description:

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Solution:

利用two pointer方法，每次对于当下要遍历的节点t1，都用进行另外一次遍历，记为t2，一直搜索到null或者t1和t2的val不同。如果t2是t1的next，那么就表示t1没有重复值，否则t1赋值到t2，继续遍历。

import java.util.*;public class Solution {public ListNode deleteDuplicates(ListNode head) {ListNode t1, t2;t1 = head;t2 = head;while (t1 != null) {t2 = t1;while (t2 != null) {if (t2.val != t1.val)break;t2 = t2.next;}if (t1.next == t2) {break;} else {t1 = t2;}}head = t1;t1 = t2 = head;ListNode temp = head;while (t1 != null) {t2 = t1;while (t2 != null) {if (t2.val != t1.val)break;t2 = t2.next;}if (t1.next == t2) {temp.next = t1;temp = temp.next;System.out.println(temp.val);t1.next = null;t1 = t2;} else {t1 = t2;}}return head;}}